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$y(x)$ is a solution of the differential equation Given that $y(x)$ is a solution of the differential equation
$$
x^2 y^{\prime \prime}+x y^{\prime}-4 y=x^2
$$
on the interval $(0, \infty)$ such that $\lim _{x \rightarrow 0^{+}} y(x)$ exists and $y(1)=1$. I need to find the value of $y^{\prime}(1)$.
$$
\begin{aligned}
& x^2 y^{\prime \prime}+x y^{\prime}-4 y=x^2 \\
& \left(x^2 D^2+x D-4\right) y=x^2 \\
& z=\log x \quad \begin{aligned}
x^2 D^2 & =D^{\prime}\left(D^{\prime}-1\right) \\
& =D^{\prime 2}-D^{\prime}
\end{aligned} \\
& \left(D^{\prime 2}-D^{\prime}+D^{\prime}-4\right) y=\left(e^z\right)^2 \\
& \left(D^{\prime 2}-4\right) y=e^{2 z} \\
& \text { Complementary Function } \\
& \\
& m^2-4=0 \\
& m^2=4 \\
& m=\pm 2 \\
& \text { c. } F=c_1 e^{2 z}+c_2 e^{-2 z} \\
& \begin{array}{l}
=c_1\left(e^z\right)^2+c_2\left(e^z\right)^{-2} \\
=c_1 x^2+c_2 x^{-2}
\end{array} \mid \\
& =c_1 x^2+\frac{1}{x^2} c_2 \\
&
\end{aligned}
$$
$$
\begin{aligned}
P \cdot I & =\frac{1}{D^{\prime 2}-4} e^{2 z} \\
& =\frac{1}{(2)^2-4} e^{2 z} \\
& =\frac{1}{4-4} e^{2 z} \quad \text { } \\
& =\frac{z}{2 D^{\prime}} e^{2 z} \\
& =\frac{z}{4} e^{2 z} \\
P I & =\frac{\log x}{4} \cdot x^2
\end{aligned}
$$
General soln,
$$
\begin{aligned}
y(x) & =C \cdot F+P \cdot I \\
& =c_1 x^2+c_2 \frac{1}{x^2}+\frac{\log x}{4} x^2
\end{aligned}
$$
Further how to get the solution ?
Thanks in advance.
|
You know that $\displaystyle\lim_{x\to 0^+}y(x)$ exists.
It is equivalent to say that, $\displaystyle\lim_{x\to 0^+}\left(c_1x^2+c_2\frac{1}{x^2}+\frac{\ln(x)}{4}x^2\right)$ exists.
You know that $\displaystyle\lim_{x\to 0^+}\frac{\ln(x)}{4}x^2=0$.
Now, you know that $\displaystyle\lim_{x\to 0^+}\left(c_1x^2+c_2\frac{1}{x^2}\right)$ exists.
The only way is that $c_2=0$
This gives: $y(x)=c_1x^2+\frac{\ln(x)}{4}x^2$
By evaluating in $x=1$, you get $y(1)=c_1$. However, $y(1)=1$. So $c_1=1$.
Finally, $y(x)=x^2+\frac{\ln(x)}{4}x^2$
I think you will be able to derive and find $y^\prime(1)$ by yourself.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4596434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Expected value to get all possible outcomes for coin flipping trials Consider the experiment where we keep flipping two fair coins. If the random variable X is defined to be the number of trials to get all possible outcomes, find E(X) and Var(X). Since the coins are distinct (H,T) and (T,H) would count as distinct outcomes.
I've been trying to think about a solution to this problem by possibly defining X in terms of 4 different random variables representing the expected number of trials for each individual outcome. I don't know how I would structure this solution however, any help would be appreciated.
|
For one coin
Let's call the number of flips as $n$.
$$
P(X=1) = 0 \\
P(X=n) = \dfrac{1}{2^{n-1}}; n = 2,3,\cdots
$$
$$
\begin{align}
\text{E}[X] &= \sum_{n=2} n\times P(n) = \sum_{n=2} \dfrac{n}{2^{n-1}} = 3 \\
\text{var}[X] &= \text{E}[X^2] - \text{E}[X]^2 \\
&= \sum_{n=2} n^2\times P(n) - 3^2 \\
&= \sum_{n=2} \dfrac{n^2}{2^{n-1}} - 9 \\
&= 2 \\
\end{align}
$$
For two coins
Let's call the number of flips of the first and second coin together as $n$.
$$
P(X = 1,2,3) = 0 \\
P(X = n) = \dfrac{3}{4^{n-3}}, n = 4,5,6,\cdots
$$
$$
\begin{align}
\text{E}[X] &= \sum_{n=4} n\times P(n) = \sum_{n=2} \dfrac{3n}{4^{n-3}} = \dfrac{13}{3} \\
\text{var}[X] &= \text{E}[X^2] - \text{E}[X]^2 \\
&= \sum_{n=4} n^2\times P(n) - \left( \dfrac{13}{3} \right) ^2 \\
&= \sum_{n=4} \dfrac{3n^2}{4^{n-3}} - \dfrac{169}{9} \\
&= \dfrac{4}{9} \\
\end{align}
$$
|
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"url": "https://math.stackexchange.com/questions/4596664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Proving $\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$ I encountered the following binomial equality:
$$\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$$
which I know it's true, but I don't know how to prove it directly. I entered the left-hand-side to Mathematica, and it directly gave me the right-hand-side. So I wonder if anyone knows an elementary proof.
|
Here is a starter. We transform the sum and separate one part which can be simplified. We start with the left hand side of OPs identity and obtain
\begin{align*}
\sum_{i=0}^K&(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}\\
&=\sum_{i=0}^K(-1)^i\frac{(2n+1-i)!}{i!(2n+1-2i)!}\,\frac{(2n-2i)!}{(K-i)!(2n-K-i)!}\\
&=\sum_{i=0}^K(-1)^i\binom{K}{i}\binom{2n-i}{K}\frac{2n+1-i}{2n+1-2i}\tag{1}\\
\end{align*}
With (1) we can reformulate OPs claim for $0\leq K\leq 2n$ as
\begin{align*}
\color{blue}{\sum_{i=0}^K(-1)^i\binom{K}{i}\binom{2n-i}{K}\frac{2n+1-i}{2n+1-2i}=\frac{1}{2}\left(1+(-1)^K\right)}\tag{2}
\end{align*}
We can write the left-hand side of (2) as
\begin{align*}
\sum_{i=0}^K(-1)^i\binom{K}{i}\binom{2n-i}{K}\left(1+\frac{i}{2n+1-2i}\right)\tag{3}
\end{align*}
In the following we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
\begin{align*}
[z^p](1+z)^q=\binom{q}{p}\tag{4}
\end{align*}
We obtain from (3)
\begin{align*}
\color{blue}{\sum_{i=0}^K}&\color{blue}{(-1)^i\binom{K}{i}\binom{2n-i}{K}}\\
&=\sum_{i=0}^K(-1)^i\binom{K}{k}\binom{2n-i}{2n-i-K}\tag{5.1}\\
&=\sum_{i=0}^K(-1)^i\binom{K}{i}[z^{2n-i-K}](1+z)^{2n-i}\tag{5.2}\\
&=[z^{2n-K}](1+z)^{2n}\sum_{i=0}^K\binom{K}{i}\left(-\frac{z}{1+z}\right)^i\tag{5.3}\\
&=[z^{2n-k}](1+z)^{2n}\left(1-\frac{z}{1+z}\right)^K\tag{5.4}\\
&=[z^{2n-k}](1+z)^{2n-K}\\
&=\binom{2n-K}{2n-K}\\
&\,\,\color{blue}{=1}\tag{5.5}
\end{align*}
Comment:
*
*In (5.1) we use $\binom{p}{q}=\binom{p}{p-q}$.
*In (5.2) we use (3).
*In (5.3) we use $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (5.4) we apply the binomial theorem and do some simplifications in the following steps.
We also obtain from (3)
\begin{align*}
\color{blue}{\sum_{i=1}^K}&\color{blue}{(-1)^i\binom{K}{i}\binom{2n-i}{K}\frac{i}{2n+1-2i}}\\
&=K\sum_{i=1}^K(-1)^i\binom{K-1}{i-1}\binom{2n-i}{K}\frac{1}{2n+1-2i}\tag{6.1}\\
&=K\sum_{i=0}^{K-1}(-1)^{i+1}\binom{K-1}{i}\binom{2n-i-1}{K}\frac{1}{2n-1-2i}\tag{6.2}\\
&=K\sum_{i=0}^{K-1}(-1)^{i+1}\binom{K-1}{i}\binom{2n-i-1}{2n-i-1-K}\frac{1}{2n-1-2i}\tag{6.3}\\
&\color{blue}{=(-1)^KK\sum_{i=0}^{K-1}\binom{K-1}{i}\binom{-K-1}{2n-i-1-K}\frac{1}{2n-1-2i}}\tag{6.4}\\
\end{align*}
Comment:
*
*In (6.1) we use $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
*In (6.2) we shift the index to start with $i=0$.
*In (6.3) we use $\binom{p}{q}=\binom{p}{p-q}$.
*In (6.4) we use $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
We finally derive from (3), (5.5) and (6.4) the following claim
\begin{align*}
\color{blue}{K\sum_{i=0}^{K-1}\binom{K-1}{i}\binom{-K-1}{2n-i-1-K}\frac{1}{2n-1-2i}=\frac{1}{2}\left(1-(-1)^K\right)}
\end{align*}
which could alternatively be used to prove OPs claim.
|
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"timestamp": "2023-03-29T00:00:00",
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|
To prove $1^1\cdot2^2\cdot 3^3...\cdot n^n<(\frac{2n+1}{3})^{\frac{n(n+1)}{2}} $ So we have to prove the following for $n\in N $ $$1^1\cdot 2^2\cdot 3^3...\cdot n^n<\left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}} $$
So I used concept of weighted means (arithmetic and geometric) used AM GM inequality.
$$AM=\frac{a_1w_1+a_2w_2+...+a_nw_n}{w_1+w_2+...+w_n}$$
$$GM=(a_1^{w_1}\cdot a_2^{w_2}\cdot...\cdot a_n^{w_n})^{\frac{1}{w_1+w_2+...+w_n}}$$
So here I let $w_1=1, w_2=2^1,w_3=3^1..$ and of course $a_1=1,a_2=2^1,a_3=3^2...$
So we get:
$$\frac{1^1+ 2^2+ 3^3...+ n^n}{\frac{n(n+1)}{2}}>(1^1\cdot 2^2\cdot 3^3...\cdot n^n)^{\frac{1}{\frac{n(n+1)}{2}}}$$
However on lhs, I cant deal with numerator, and I feel that if it can be simplified, I would get the answer. So please help or if possible suggest new method.
|
Use the AM-GM inequality [where there are $\frac{n(n+1)}{2}$ terms; indeed for each positive integer $i \le n$, there are $i$ terms of $i$]:
$$\frac{\sum_{1=1}^n i^2}{n(n+1)/2} \ \ge \ \sqrt[\frac{n(n+1)}{2}]{1^12^2 \cdots n^n},$$
where in the above inequality, the LHS represents the arithmetic mean of the above terms and the RHS the geometric mean of the above terms.
However, the equation
$$\frac{\sum_{1=1}^n i^2}{n(n+1)/2} = \frac{(n)(n+1)(2n+1)}{6} \times \frac{2}{n(n+1)}$$
$$ = \frac{2n+1}{3} $$ also holds. So then combining this string of equations with the top AM-GM inequality, yields the inequality
$$\frac{2n+1}{3} \ \ge \ \sqrt[\frac{n(n+1)}{2}]{1^12^2 \cdots n^n} \ .$$
Raising each side of this to the $\frac{n(n+1)}{2}$-power yields the desired result.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4598722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
If $\frac{\sin2x}{\cos x}-1=0$, find $x$ given $\ \frac{\pi}{2}\le x\le\pi$ (cancelling out vs multiplying out the $\cos x$) To solve the equation
$$\frac{\sin2x}{\cos x}-1=0 \tag1$$
I manipulated it into the following:
$$\frac{2\sin x\cos x}{\cos x}-1=0 \tag2$$
Then, by multiplying $\cos x$ out, I obtained:
$$2\sin x\cos x-\cos x=0 \tag3$$
However, according to the answers I was supposed to cancel out the $\cos x$ which results in
$$2\sin x-1=0 \tag4$$ Thus, I myself attained the answers $\frac{5\pi}{6},\ \frac{\pi}{2}$, whilst the answer booklet merely had $\frac{5\pi}{6}$ .
I do not understand the logic behind cancelling out the $\cos x$ - I thought we were supposed to maximise the number of answers obtained by multiplying out rather than cancelling.
Any help clearing up my confusion would be much appreciated.
|
The confusion is due to the violation of equivalence conditions between mathematical steps.
Problem statement:
$$\frac{\sin2x}{\cos x}-1=0$$
where, $\frac {\pi}{2}\leqslant x\leqslant\pi$.
$\rm Step-1.$
$$\frac{2\sin x\cos x}{\cos x}-1=0$$
There is no issue here. Because: $\sin 2x=2\sin x\cos x$.
$\rm Step-2.$
$$2\sin x\cos x-\cos x=0$$
We want to analyze this step. When we multiply both side of the equation by $\cos x$, we should do this under the restriction $\cos x≠0$. So why?
*
*No matter what mathematical operation we apply, before we start solving the equation $\frac{\sin2x}{\cos x}-1=0$, it is necessary to state that $\cos x≠0$. Because, we can never make the denominator $0$.
*Applying the multiplication we have:
$$\cos x \left(\frac{2\sin x\cos x}{\cos x}-1\right)=0\cdot \cos x$$
This is essentially equivalent to:
$$\cos x \cdot \frac{2\sin x\cos x}{\cos x}-\cos x=0$$
Then, we can easily reduce the last equation to the following equivalent equation:
$$\begin{align}\frac {\cos x}{\cos x} \cdot \left(2\sin x\cos x\right)-\cos x=0\end{align}$$
Think about: if $\cos x=0$, can we write $\frac {\cos x}{\cos x}=\frac 00=1$ ?
Since $\cos x$ can not equal to $0$, then the multiplication under the restriction $\cos x≠0$, (even if $\cos x$ is not in the denominator) doesn't produce roots that are called extraneous roots in mathematics.
Therefore, the original equation $\frac{\sin2x}{\cos x}-1=0$ is equivalent to the derived equation $2\sin x\cos x-\cos x=0$, if and only if, when $\cos x≠0$.
The possible correct solution can be reached with the following $2$ ways:
$\require {cancel}$
$$\begin{align}&\frac{\sin2x}{\cos x}-1=0\\
\iff &\frac {2\sin x\cdot\cancel{ {\color{#c00}{\cos x}}}}{\cancel{\color{#c00}{{\cos x}}}}-1=0\\
\iff &2\sin x-1=0,\thinspace \cos x≠0\\
\iff &\sin x=\frac 12,\thinspace\cos x≠0\end{align}$$
or
$$\begin{align}&\frac{\sin2x}{\cos x}-1=0\\
\iff &\frac {2\sin x\cos x}{\cos x}-1=0\\
\iff &\frac {2\sin x\cos x-\cos x}{\cos x}=0.\end{align}$$
Then we know that, $$\frac {A(x)}{B(x)}=0\iff A(x)=0\wedge B(x)≠0.$$
Therefore, we have:
$$2\sin x\cos x-\cos x=0,\thinspace\cos x≠0$$
This leads to:
$$\begin{align}&\cos x\left( 2\sin x-1\right)=0,\thinspace\cos x≠0\\
\iff &2\sin x-1=0,\thinspace\cos x≠0\\
\iff &\sin x=\frac 12,\thinspace \cos x≠0.\end{align}$$
Finally, under the restriction $\frac {\pi}{2}\leqslant x\leqslant\pi$, we obtain:
$$\sin x=\frac 12\wedge x≠\frac {\pi}{2}$$
Thus, indeed the answer is $x=\frac {5\pi}{6}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$f(x)=\sqrt[3]{x^2}$ derivative by defintion
Find $f(x)=\sqrt[3]{x^2}$ derivative by definition
my first try was
$\lim_\limits{h \to 0}\frac{\sqrt[3]{(x+h)^2}-\sqrt[3]{x^2}}{h}$
I tried multiplying by $\frac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2}}$ and got$\lim_\limits{h \to 0}\frac{(x+h)(\sqrt[3]{x+h})-x\sqrt[3]x}{h\cdot(\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2})}$
but I don't know how to continue from here
I also tried multiplying by $\frac{(x+h)^{1/3}+x^{1/3}}{(x+h)^{1/3}+x^{1/3}}$ but this seemed more confusing
Also tried writing $\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2}$ as $(\sqrt[3]{(x+h)}-\sqrt[3]x)\cdot(\sqrt[3]{x+h}+\sqrt[3]x$ but this also did not help..
thanks for any help and tips
|
First of all, I would suggest you to apply the following identity:
\begin{align*}
a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})
\end{align*}
According to it, one concludes the desired result:
\begin{align*}
f'(x_{0}) & = \lim_{x\to x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}}\\
& = \lim_{x\to x_{0}}\frac{x^{2/3} - x^{2/3}_{0}}{x - x_{0}}\\
& = \lim_{x\to x_{0}}\frac{x^{2} - x^{2}_{0}}{(x - x_{0})(x^{4/3} + x^{2/3}x^{2/3}_{0} + x^{4/3}_{0})}\\
& = \lim_{x\to x_{0}}\frac{x + x_{0}}{x^{4/3} + x^{2/3}x^{2/3}_{0} + x^{4/3}_{0}}\\
& = \frac{2x_{0}}{3x^{4/3}_{0}} = \frac{2}{3x^{1/3}_{0}}
\end{align*}
Hopefully this helps!
|
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|
A pen-and-paper proof for a matrix implication. Suppose $A = \begin{bmatrix} x & 1\\ y & 0\end{bmatrix}, B = \begin{bmatrix} z & 1\\ w & 0\end{bmatrix}$, for $x,y,z,w \in \Bbb{R}$.
I have observed by considering many examples of $x,y,z,w$ that:
If all the eigen values of $A^2B$ and $AB^2$ are less than one in absolute value $\implies$ $\det(AB+A+I)<0$ and $\det(BA+B+I)<0$ is not possible.
OR alternatively,
If all the eigen values of $A^2B$ and $AB^2$ are less than one in absolute value $\implies$ $\det(AB+A+I)\ge 0$ OR $\det(BA+B+I)\ge 0$
I wonder how to prove it actually?
A computational proof using computer package was shown in
https://mathoverflow.net/questions/435267/proof-of-a-matrix-implication/435689#435689
But I am wondering about formal or analytical proof for this question which can be done using pen paper.
$\textbf{EDIT}$
$\textbf{The case of $y=x$ or $z=w$ is covered by Andreas as an answer below.}$
$\textbf{The only case that remains to show is whether the conjecture still holds for $y\neq x$ or $w \neq z$.}$
$\textbf{The bounty is for this purpose if some has ideas on it?}$
|
The claim in the question can be rephrased by taking the inverse formulation:
$\det(AB+A+I) < 0 \qquad $ AND $\qquad \det(BA+B+I) < 0$
$\implies$ At least one eigenvalue of $(A^2B$ ; $AB^2)$ is greater or equal than one in absolute value.
As a boundary case, there are indeed situations with
$\det(AB+A+I) = 0$ ; $\det(BA+B+I)= 0 $ ;
where at least one eigenvalue of $(A^2B$ ; $AB^2)$ equals $1 +q$ where $q > 0$.
This is e.g. achieved with values $x = w = z = -1 $ and $y = -1-q$, then we have
$$
\det(AB+A+I) = (x-y)z+(y+1)w+(x+1) = 0\\
\det(BA+B+I)=(z-w)x+(w+1)y+(z+1) = 0
$$
and the characteristic equations for the eigenvalues read
$$
\det(A^2 B-\lambda I) = (z(x^2+y)+xw - \lambda)(x y - \lambda) - (x^2+y)(xyz+wy) = (1+q -\lambda )^2 = 0 \\
\det(A B^2-\lambda I) = (x(w+z^2)+wz - \lambda) (y z - \lambda) - (xyz+wy)(z^2+w) = (1+q -\lambda )(1 -\lambda )= 0
$$
Hence we have eigenvalues $\lambda = 1+q > 1$ for both $A^2 B$ and $A B^2$.
This could also support possible contradictions to the claim, however not sufficiently. It gives rise to be able, by a continuity argument, to transport this to the adjacent domain $\det(AB+A+I) < 0 \qquad $ AND $\qquad \det(BA+B+I) < 0$. We can look at this domain by observing
$$
\det(AB+A+I) = (x-y)z+(y+1)w+(x+1) = (y-x)(w-z) + (x+1)(w+1) \tag{1}\\
\det(BA+B+I)=(z-w)x+(w+1)y+(z+1) = (y-x)(w-z) + (y+1)(z+1)
$$
Hence a particularly convenient subcase is the one with $y=x$ and $w=z$. Here, $\det(AB+A+I) < 0$ is achieved in the two cases $(x>-1 ; z < -1)$ and $(x<-1 ; z > -1)$. It is only necessary to consider one of these cases, since exchanging $x \leftrightarrow z$ exchanges $A \leftrightarrow B$ and this is covered by the second determinant inequality. As we are only aiming at showing that one eigenvalue will always be greater or equal than one in absolute value, it suffices to establish one case.
We have the characteristic equation for $A^2 B$:
\begin{align}
& \det(A^2 B - \lambda I) =
\lambda^2 - \lambda(x z^2 + z^2 + 2 x z) - x z^2 = 0
\end{align}
Now let $x = -1 +a$ and $z = -1 -b$ with $a,b > 0$, then we get
$$
(\lambda-1)^2 + \lambda(-a b^2- 2 b +a) -(a - 2 b + 2a b + b^2 a - b^2) = 0
$$
which can be solved for $\lambda$. To simplify things, it suffices to notice that the equation
$$
(\lambda-1)^2 + \lambda q - (q + r) = 0
$$
has the largest solution
$$
\lambda_1 = 1 - q/2 + \frac12 \sqrt{q^2 + 4r}
$$
Hence we have, if $r >0$, that $\lambda_1 > 1 + (1 - {\rm sign}(q)) \; |q|/2 > 1$. Applied to the characteristic equation, this is
$$
r = (a - 2 b + 2a b + b^2 a - b^2)- (-a b^2- 2 b +a) = b(2 a b + 2 a - b)
$$
hence if $2 a b + 2 a - b > 0$ then $A^2 B$ has an eigenvalue $\lambda > 1$.
Likewise, the characteristic equation for $A B^2$ gives that if $2b - a - 2ab>0$, then $A B^2$ has an eigenvalue $\lambda > 1$.
Adding the two conditions gives $a+b$ which is $> 0$, hence it is impossible that both conditions fail (i.e. both are negative). Hence, at least one eigenvalue of $(A^2B$ ; $AB^2)$ is greater or equal than one in absolute value.
This establishes the claim for the subcase $y=x$ and $w=z$. The next step is to exploit the determinant equations (1) further for the case $y\ne x$ or $w\ne z$. [...]
|
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|
Proving the a matrix is bounded for every power I want to show that the following matrix satisfies $\|A^n(z)\| \leq C$ for every $z \in \mathbb{C}$ with $Re(z) \leq 0$, for $n= 0,1,2,...$, and some constant $C$. How do I approach this kind of problems?
$$A(z) = \begin{bmatrix} \frac{z^2+6z+8}{2(2-z)^2} & -\frac{z^2+10z+24}{8(2-z)^2} \\ \frac{2z(z+2)}{(2-z)^2} & -\frac{z(z+6)}{2(2-z)^2} \end{bmatrix}$$
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Given matrix $M=\{M_{ij}\}$ and $N=\{N_{ij}\}$ where $M_{ij}\in\Bbb C$ and $N_{ij}\in\Bbb R_{>0}$, write $M\prec N$ if $|M_{ij}|<N_{ij}$ for all $i,j$. It is straightforward to verify that if $M\prec N$ and $M'\prec N'$, then $MM'\prec NN'$.
It is enough to prove the following claim.
Claim: Let $z \in \mathbb{C}$ with $\Re(z) \leq 0$. Then $A^n(z) \prec \begin{bmatrix} \frac12& \frac18 \\ 2 & \frac12\end{bmatrix}$ for all $n=1,2,3,\cdots$.
Proof:
Since $\Re(z)<0$, we have $|z+2|<|2-z|$, $\ |z+4|<|2-z|$ and $|z+6|<|2-z|$. $$A(z) = \begin{bmatrix} \frac{(z+2)(z+4)}{2(2-z)^2} & -\frac{(z+4)(z+6)}{8(2-z)^2} \\ \frac{2z(z+2)}{(2-z)^2} & -\frac{z(z+6)}{2(2-z)^2} \end{bmatrix}\prec \begin{bmatrix} \frac12& \frac18 \\ 2 & \frac12\end{bmatrix},$$ which means the claim is true for $n=1$.
Suppose the claim is true for $n$, i.e., $A^n(z)\prec \begin{bmatrix} \frac12& \frac18 \\ 2 & \frac12\end{bmatrix}$. Then
$$A^{n+1}(z)=A^n(z)A(z)\prec \begin{bmatrix} \frac12& \frac18 \\ 2 & \frac12\end{bmatrix}\begin{bmatrix} \frac12& \frac18 \\ 2 & \frac12\end{bmatrix}=\begin{bmatrix} \frac12& \frac18 \\ 2 & \frac12\end{bmatrix}.$$ Hence, the claim is true for $n+1$, too. $\quad\checkmark$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4601971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Assuming x is small, expand $\frac{\sqrt{1-x}}{\sqrt{1+2x}}$ up to and including the term in $x^{2}$ I have tried this many times but can't quite land on the correct answer.
The correct answer:
$1-\frac{3x}{2}+\frac{15x^{2}}{8}$
These are the steps I took:
*
*Re wrote it as: $\left ( 1-x \right )^{\frac{1}{2}}\left ( \left ( 1+2x \right )^{\frac{1}{2}} \right )^{-1}$
*Using bionmal expansion I expanded $\left ( 1-x \right )^{\frac{1}{2}}$ up to $x^{2}$
*
*For this I get : $\left (1-\frac{x}{2}-\frac{x^{2}}{8} \right)$
*Then to solve the second bracket written in step 1 I did $\left (1+2x \right )^{\frac{1}{2}}$
*
*For this I got $\left (1-x+\frac{x^{2}}{4} \right)$
*Then finally I raised this to the power one so $\left (1-x+\frac{x^{2}}{4} \right)^{-1}$
*For this I got $\left (1+x+\frac{3x^{2}}{4} \right)$
*Then I just expaned those two brakcets so: $\left (1-\frac{x}{2}-\frac{x^{2}}{8} \right)$$\left (1+x+\frac{3x^{2}}{4} \right)$
Once expanded, I then neglected powers bigger than $x^{2}$ (mentioned in question). Then collected like terms. However either my method or the algebra is going wrong, and I just need some help with this.
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$$\frac{\sqrt{1-x}}{\sqrt{1+2x}}=(1-x)^{1/2}(1+2x)^{-1/2}$$
Consider that $(1+t)^{\alpha}\sim 1+\alpha t+\frac{1}{2}(\alpha -1)\alpha t^2$, hence you obtain:
$$ (1-x)^{1/2}\sim 1-\frac{x}{2}-\frac{x^2}{8} \quad \text{and}\quad (1+2x)^{-1/2}\sim 1-x+\frac{3x^2}{2}$$
It follows that:
$$\left(1-\frac{x}{2}-\frac{x^2}{8}\right)\left(1-x+\frac{3x^2}{2}\right)\sim 1-x+\frac{3x^2}{2}-\frac{x}{2}+\frac{x^2}{2}-\frac{x^2}{8}=1-\frac{3x}{2}+\frac{15x^2}{8}$$
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating a double integral I was trying to evaluate the following integral
$$\int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{y \ln y \ln x}{(x^2+ y^2)( 1+y^2)} dy dx.$$
I have a guess that the value of this integral is $\frac{\pi^4}{8}$. But I am unable to prove it.
Could someone please help me in evaluating this integral?
Or, can we show the following identity holds without much calculation?
$$\int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{2y \ln y \ln x}{(x^2+ y^2)( 1+y^2)} dy dx= \int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{y (\ln y )^2}{(x^2+ y^2)( 1+y^2)} dy dx.$$
Any help or hint would be appreciated. Thanks in advance.
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$$I=\int\frac{y \log(y) }{(x^2+ y^2)( 1+y^2)} dy $$ is not bad since
$$A=\frac{y }{(x^2+ y^2)( 1+y^2)}=\frac{y}{(y-i) (y+i) (y-i x) (y+i x)}$$ Using partial fraction decomposition
$$A=\frac 1{2(x^2-1)}\left(\frac 1{y-i} +\frac 1{y+i}-\frac 1{y-ix}-\frac 1{y+ix}\right)$$ and
$$\int \frac {\log(y)}{y+a}\,dy=\text{Li}_2\left(-\frac{y}{a}\right)+\log (y) \log \left(1+\frac{y}{a}\right)$$ Therefore
$$I=\frac 1{2(x^2-1)}\left(\frac{1}{2}
\left(\text{Li}_2\left(-y^2\right)-\text{Li}_2\left(-\frac{y^2
}{x^2}\right)\right)+\log (y) \log \left(\frac{x^2
\left(y^2+1\right)}{x^2+y^2}\right) \right)$$ and then
$$J=\int_0^\infty\frac{y \log(y) }{(x^2+ y^2)( 1+y^2)} dy =\frac{\log ^2(x)}{2 \left(x^2-1\right)}$$
Then, what remains is the computation of
$$K=\int \frac{\log ^3(x)}{x^2-1}\,dx=\frac 12\left(\int \frac{\log ^3(x)}{x-1}\,dx- \int \frac{\log ^3(x)}{x+1}\,dx\right)$$ which is not too bad using a few integrations by parts.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
How to find the mass of a paraboloid region underneath a plane given a non-constant density Given the paraboloid $z=4x^2 + 4y^2$, the plane $z=4$, and the density function $\sigma(x, y, z) = z + 2y$, find the mass of the region bounded by the paraboloid under the plane.
Intuitively, this makes the most sense using cylindrical coordinates. And since mass is just volume multipied by density (?), here's my (incorrect) attempt:
\begin{align}
z = & \; 4x^2 + 4y^2 \implies z = 4r^2 \\
\implies& \int_0^{2\pi}\int_0^1\int^4_{4r^2} \sigma(x, y, z) \, \cdot \, rdzdrd\theta = \int_0^{2\pi}\int_0^1\int^4_{4r^2} (4r^3+2r^2\sin\theta) \; dzdrd\theta \\
= \ & \int_0^{2\pi}\int_0^1 (4r^3+2r^2\sin\theta)(4-4r^2) \; drd\theta \\
= \ & 4\int_0^{2\pi}\int_0^1 (4r^3 + 2r^2\sin\theta - 4r^5 - 2r^4\sin\theta) \; drd\theta \\
= \ & 4\int_0^{2\pi} (\frac{1}{3} + \frac{2\sin\theta}{3} - \frac{2\sin\theta}{5}) \; d\theta = 4(2\pi/3) = 8\pi/3\\
\end{align}
However, the correct answer is double of that, $16\pi/3$.
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You almost got it right. It should be$$\int_0^{2\pi}\int_0^1\int_{4r^2}^4\bigl(\color{red}{zr}+2r^2\sin(\theta)\bigr)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{16\pi}3,$$since $\sigma(r\cos(\theta),r\sin(\theta),z)=z+2r\sin(\theta)$.
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"url": "https://math.stackexchange.com/questions/4607321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $\int_0^{\frac{\pi}{2}}x\ln\tan xdx = \sum_{n=0}^{+\infty}\frac{1}{(2n+1)^3}=\frac78\zeta(3)$. Prove that $\int_0^{\frac{\pi}{2}}x\ln\tan xdx = \sum_{n=0}^{+\infty}\frac{1}{(2n+1)^3}=\frac78\zeta(3)$.
I think it's natural to use series, but if I take $u=\tan x$ the integral becomes: $\int_{0}^{+\infty}\frac{\arctan u\ln u}{u^2+1}du$.
If I expand $f(x)=\frac{\arctan u}{u^2+1}$ to series, the result is not convergent. However, the answer is so simple! Is there really a direct method to solve it?
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Let's consider a more general case and denote
$\displaystyle I(\alpha)=\int_0^\frac{\pi}{2}\tan^\alpha (x)\,x\,dx=\int_0^\infty t^\alpha\arctan t\,\frac{dt}{1+t^2};\,\alpha\in(-2;1)\tag*{}$
Then
$$I_k=\int_0^{\frac{\pi}{2}}x\ln^k(\tan x)dx=\int_0^\infty \ln^kt\,\arctan t\,\frac{dt}{1+t^2}=\frac{d^k}{d\alpha^k}I(\alpha)\,\bigg|_{\alpha=0}, \,k=0,1, 2...\tag{0}$$
Using $\, \arctan t=\int_0^1\frac{t\,dx}{1+(xt)^2}$, we can write $I(\alpha)$ in the form
$\displaystyle I(\alpha)=\int_0^1\frac{dx}{x^2}\int_0^\infty\frac{t^{\alpha+1}}{(t^2+1)\big(t^2+\frac{1}{x^2}\big)}dt=\int_0^1\frac{dx}{x^2}J(\alpha, x)\tag*{(1)}$
Using complex integration (along the keyhole contour)
$\displaystyle \Big(1-e^{2\pi i\alpha}\Big)J(\alpha, x)=2\pi i\underset{z=\pm i; \,\pm\frac{i}{x}}{\operatorname{Res}}\frac{z^{\alpha+1}}{(z^2+1)\big(z^2+\frac{1}{x^2}\big)}\tag*{}$
The residues evaluation is straightforward. We get
$\displaystyle J(\alpha, x)=\frac{i\,x^2}{1-x^2}\Big(1-\frac{1}{x^\alpha}\Big)\frac{e^\frac{\pi i\alpha}{2}+e^\frac{3\pi i\alpha}{2}}{1-e^{2\pi i\alpha}}=\frac{\pi}{2\sin\frac{\pi\alpha}{2}}\frac{x^2}{1-x^2}\frac{1-x^\alpha}{x^\alpha}\tag*{(2)}$
Putting (2) into (1)
$\displaystyle I(\alpha)=\frac{\pi}{2\sin\frac{\pi\alpha}{2}}\int_0^1\frac{1-x^\alpha}{x^\alpha}\frac{dx}{1-x^2}=\frac{\pi}{4\sin\frac{\pi\alpha}{2}}\int_0^1\Big(t^{-\frac{1+\alpha}{2}}-t^{-\frac{1}{2}}\Big)\frac{dt}{1-t}\tag*{}$
To perform integration, we use regularization and the main presentation of beta-function
$\displaystyle I(\alpha)=\lim_{\epsilon\to 0}\frac{\pi}{4\sin\frac{\pi\alpha}{2}}\int_0^1\Big(t^{-\frac{1+\alpha}{2}}-t^{-\frac{1}{2}}\Big)(1-t)^{\epsilon-1}dt\tag*{}$
$\displaystyle =\lim_{\epsilon\to 0}\frac{\pi}{4\sin\frac{\pi\alpha}{2}}\Gamma(\epsilon)\bigg(\frac{\Gamma\big(\frac{1-\alpha}{2}\big)}{\Gamma\big(\frac{1-\alpha}{2}+\epsilon\big)}-\frac{\Gamma\big(\frac{1}{2}\big)}{\Gamma\big(\frac{1}{2}+\epsilon\big)}\bigg)\tag*{}$
Denoting $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ - digamma-function, we get the answer:
$\displaystyle I(\alpha)=\frac{\pi}{4}\,\frac{\psi\big(\frac{1}{2}\big)-\psi\big(\frac{1-\alpha}{2}\big)}{\sin\frac{\pi\alpha}{2}}\tag*{(3)}$
$\displaystyle \boxed{\,I_k=\frac{d^k}{d\alpha^k}I(\alpha)\,\bigg|_{\alpha=0}=\frac{\pi}{4}\,\frac{d^k}{d\alpha^k}\,\bigg|_{\alpha=0}\frac{\psi\big(\frac{1}{2}\big)-\psi\big(\frac{1-\alpha}{2}\big)}{\sin\frac{\pi\alpha}{2}}\,\,}\tag*{(4)}$
Now, disclosing the uncertainty,
$\displaystyle I_0=I(0)=\lim_{\alpha\to 0}\frac{\pi}{4}\frac{\psi\big(\frac{1}{2}\big)-\psi\big(\frac{1}{2}\big)+\psi^{(1)}\big(\frac{1}{2}\big)\frac{\alpha}{2}+O(\alpha^2)}{\sin\frac{\pi\alpha}{2}}=\frac{\psi^{(1)}\big(\frac{1}{2}\big)}{4}=\frac{\pi^2}{8}\tag*{}$
as it should be.
The desired integral $I_1$
$$I_1=\frac{\pi}{4}\,\frac{d}{d\alpha}\,\bigg|_{\alpha=0}\frac{\psi\big(\frac{1}{2}\big)-\psi\big(\frac{1-\alpha}{2}\big)}{\sin\frac{\pi\alpha}{2}}=-\,\frac{\psi^{(2)}\big(\frac{1}{2}\big)}{8}=\sum_{k=1}^\infty\frac{1}{(2k-1)^3}=\frac{7}{8}\zeta(3)$$
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"url": "https://math.stackexchange.com/questions/4609446",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after this I was unable to make any further progress in deducing the desired inequality.
I appreciate your help
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$$x^4+3x^3+9x^2+3x+1=\left(x^2+\frac32x+1\right)^2+\frac{19}4x^2$$ is always positive.
There is a general theorem that if a real polynomial is always non-negative, it can be written as a sum of squares.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4611737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Proving $\left\lfloor(\frac{1+\sqrt{5}}{2})^{4n+2}\right\rfloor-1$ is a perfect square for $n=0,1,2,\ldots$ Let $$S_n = \left \lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1$$ ($n=0, 1, 2, \ldots$).
Prove that $S_n$ is a perfect square.
In Art of Problem Solving website, there is a hint
$$
\begin{align}
\left\lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1 & =\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}+\left(\frac{1-\sqrt{5}}{2}\right)^{4n+2}-2\\
&=\left(\left(\frac{1+\sqrt{5}}{2}\right)^{2n+1}+\left(\frac{1-\sqrt{5}}{2}\right)^{2n+1}\right)^2
\end{align}
$$
I don't know how to get the first equal sign, is it mean
$$
\left\lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor = \left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}-\left\{\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\} $$
But how to prove the decimal part of $\phi^{4n+2}$
$$ \left\{\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2} \right\} =1-\left(\frac{1-\sqrt{5}}{2}\right)^{4n+2}$$
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Let $$X_n=(\frac{1+\sqrt{5}}{2})^{4n-2}=(\frac{3+\sqrt{5}}{2})^{2n-1}=I+f, I \in I^+, 0 < f <1,0 <f'=(\frac{3-\sqrt{5}}{2})^{2n-1}<1$$
$$X_n+f'=I+f+f'
=(\frac{3+\sqrt{5}}{2})^{2n-1}+(\frac{3-\sqrt{5}}{2})^{2n-1}$$
$$= (7+3\sqrt{5})^{n}\frac{2}{3+\sqrt{5}}+(7-3\sqrt{5})^{n}\frac{2}{3-\sqrt{5}}=3A_n-5B_n=K$$
Where $(7+3\sqrt{5})^n=A_n+B_n\sqrt{5}, A_n. B_n \in I^+$
$K$ is integer,so is $I+f+f'$. This implies that $f+f'$ is also an integer as $0<f+f'<2$ and it equals 1. Hence $I=K-1.$
Check that $$K=\left((\frac{1+\sqrt{5}}{2})^{2n-1}+(\frac{1-\sqrt{5}}{2})^{2n-1}\right)^2+2. $$
$$\implies I-1=\left((\frac{1+\sqrt{5}}{2})^{2n-1}+(\frac{1-\sqrt{5}}{2})^{2n-1}\right)^{2}.$$
Check that $$(\frac{1+\sqrt{5}}{2})^{2n-1}+(\frac{1-\sqrt{5}}{2})^{2n-1}=(3+\sqrt{5})^n\frac{2}{1+\sqrt{5}}+(3-\sqrt{5})^n\frac{2}{1-\sqrt{5}}$$
$$=5D_n-C_n,~ \text{where} ~(3+\sqrt{5})^n=C_n+D_n\sqrt{5}, C_n, D_n \in I^{+}.$$
Finally we have $I-1=(5D_n-C_n)^2, C_n, D_n \in I^+$
|
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"url": "https://math.stackexchange.com/questions/4620233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Solving $y''(x) + \epsilon y'(x) + 1 = 0$ using power series We are given
\begin{align}
\begin{cases}
y''(x) + \epsilon y'(x) + 1 =0, \ 0 < \epsilon <<1\\
y(0)=0, \ y'(0)=1
\end{cases}
\end{align}
and asked to solve this using the solution form $y(x) = \sum_{n=0}^{\infty}\epsilon^{n}y_n(x)$.
Doing the known method for power series ODEs, we have
\begin{align}
&y(x) = \sum_{n=0}^{\infty}y_n(x) \epsilon^n\\
&y'(x) = \sum_{n=1}^{\infty}ny_n(x) \epsilon^{n-1} = \sum_{n=0}^{\infty}(n+1)y_{n+1}(x)\epsilon^{n}\\
&y''(x) = \sum_{n=2}^{\infty}n(n-1)y_n(x) \epsilon^{n-2} = \sum_{n=0}^{\infty}(n+2)(n+1)y_{n+2}(x)\epsilon^{n}.
\end{align}
So we must plug them into the system and find the general type $y_n$. I know how to do this in general. However, the present systems grinds to a halt, at least to my eyes.
Doing the substitution
\begin{align}
&\sum_{n=0}^{\infty} (n+2)(n+1)y_{n+2}(x)\epsilon^n + \sum_{n=0}^{\infty}(n+1)y_{n+1}(x) \epsilon^{n+1} + 1 =0\\
&1+ \sum_{n=0}^{\infty}(n+1)(n+1)y_{n+2}(x) \epsilon^{n} + \sum_{n=1}^{\infty}ny_n(x)\epsilon^{n} =0\\
&1 + 2y_2(x) + \sum_{n=1}^{\infty}\left\{ (n+2)(n+1)y_{n+2}(x) + ny_{n}(x) \right\}\epsilon^{n} = 0
\end{align}
and this means that
\begin{align}
\begin{cases}
y_2 = -\dfrac{1}{2}\\
y_{n+2} = \dfrac{-n}{(n+2)(n+1)}y_n
\end{cases}
\end{align}
but I think that leads to an algebraic fault, since by plugging $n=0$ at the second one we get $y_2(x) = 0$, but we found out that $y_2(x) = -\dfrac{1}{2}$.
Any thoughts on how to proceed?
EDIT: The answer is hinted to be
\begin{align}
y_{n}(x) = (-1)^{n} \left[ \dfrac{x^{n+1}}{(n+1)!} - \dfrac{x^{n+2}}{(n+2)!} \right]
\end{align}
|
One can solve directly
$$
y''+(ϵy'+1)=0\implies ϵy'(x)+1=(ϵy'(0)+1)e^{-ϵx}=(1+ϵ)e^{-ϵx}
\\
\implies
ϵy(x)+x=(1+ϵ)\frac{1-e^{-ϵx}}{ϵ}=(1+ϵ)\left(x-ϵ\frac{x^2}{2}+ϵ^2\frac{x^3}{3!}-ϵ^3\frac{x^4}{4!}+\dots\right)
\\
y(x)=x-\frac{x^2}{2}-ϵ\left(\frac{x^2}{2}-\frac{x^3}{3!}\right)+ϵ^2\left(\frac{x^3}{3!}-\frac{x^4}{4!}\right)+\dots
$$
So indeed, the reference solution
$$y(x)=\sum_{n=0}^\infty ϵ^ny_n(x),~~~y_n(x)=(-1)^n\left(\frac{x^{n+1}}{(n+1)!}-\frac{x^{n+2}}{(n+2)!}\right)
$$
is correct (contrary to my comment).
The x-derivatives of the perturbation series are, under the assumption of uniform convergence of all related series,
$$
y'(x)=\sum_{n=0}^\infty ϵ^ny_n'(x),~~~ y''(x)=\sum_{n=0}^\infty ϵ^ny_n''(x)
$$
After comparing coefficients, the resulting system is
$$
y_0''(x)+1=0,~~y_0(0)=0,~y_0'(0)=1\\
y_n''(x)+y_{n-1}'(x)=0,~~y_n(0)=y_n'(0)=0,~~n>0\\
$$
which indeed gives the same functions as solutions.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the real part of $\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}$?
What is the real part of this complex number?
$$\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}$$
I am trying to times conjugate of denominator which will be $$\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}\cdot\frac{e^{-i\theta}-1}{e^{-i\theta}-1}$$
but it makes my fractions very complicated and I don't know how do I go on.
Thanks for any help.
|
The real part of $\large\frac{a+bi}{c+di}$ is $\large\frac{ac+bd}{c^2+d^2}$ so the real part of $\large\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}=\frac{\cos(n+1)\theta-1+i\sin(n+1)\theta}{\cos\theta-1+i\sin\theta}$ is
$$\frac{(\cos(n+1)\theta-1)(\cos\theta-1)+\sin(n+1)\theta\sin\theta}{(\cos\theta-1)^2+(\sin\theta)^2}\\
=\frac{(-2)\sin^2\frac{(n+1)\theta}{2}(-2)\sin^2\frac{\theta}{2}+2\sin\frac{(n+1)\theta}{2}\cos\frac{(n+1)\theta}{2}.2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2-2\cos\theta}\\
=\frac{4\sin\frac{(n+1)\theta}{2}\sin\frac{\theta}{2}\left(\sin\frac{(n+1)\theta}{2}\sin\frac{\theta}{2}+\cos\frac{(n+1)\theta}{2}\cos\frac{\theta}{2}\right)}{4\sin^2\frac{\theta}{2}}\\
=\frac{\sin\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\\
=\sin\frac{(n+1)\theta}{2}\cos\frac{n\theta}{2}\csc\frac{\theta}{2}$$
I don't recommend this way. It took my life. I will not check again.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determining the sum of $\frac{a_{n+1}}{a_n}$ where $a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$
Let $a_0=1, a_1=\frac{1}{2}, a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$, then find $\lim_{n\to\infty} \sum_{k=0}^{n}\frac{a_{k+1}}{a_k}=\dots$
We have $(a_n)$ is strictly decreasing as $a_{n+1}-a_n=\frac{-a_n(a_n+1)}{1+(n+1)a_n}<0, a_n>0$ and $ |{\frac {a_{n+1}}{a_n}|}<1$, then the sum is convergent.
Tried to telescope the sum by defining $\frac{a_{n+1}}{a_n}=\frac{na_n}{1+(n+1)a_n}=b_n$ (say). Eventually there was coming another sequence that was not giving any conclusion. Tried examining the behaviour of the terms of this sequence $$ a_0=1,a_1=\frac{1}{2},a_2=\frac{1}{2^3},a_3=\frac{1}{2^2(2^3+2+1)}\\a_4=\frac{3^2}{2^4(2^3+2^2)(2^3+3)(2^4(2^3+2^2)(2^3+2^2+1)+2^3+2^2+2+1)}$$
There would come up more $2^n$'s in the denominator for larger n's. Tried mimicking the solution as in the Convergence of $\left(a_{n+1}= \cfrac{{a_n}^2}{1+{a_n}} (n\ge 1) , a_1=1 \right)$ (seems a bit relatable), but I am stuck here.
Any hint would be appreciated. Thanks.
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You can rearrange your equation in the following manner
$a_{n+1} (1 + (n+1) a_n) = n a_n^2$
$ na_n^2-(n+1)a_{n+1}a_n = a_{n+1}$
$\frac{a_{n+1}}{a_n} = na_n-(n+1)a_{n+1}$
The terms on the right hand side cancel out once you apply summation on both sides leaving
$\Sigma_{n=1}^{\infty} \frac{a_{n+1}}{a_n} = a_1$.
Note: Let the limit of $a_n$ as n approaches $\infty$ be $L_\infty$. You get two values for the limit: 0 and -1 by using the second equation above.
$L_\infty^2 n -(n+1)L_\infty^2= L_\infty$
$-L_\infty^2 = L_\infty$
Since all the terms are positive as you have shown $L_\infty = 0$.
Edit (Thanks to @Matija for pointing this out):
The convergence of the series $na_n$ needs to be shown instead of the series $a_n$.
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|
Probability of a sum being divisible by $4$ Roll $n$ six-sided dice.
What is the probability that the sum of the results is divisible by $4$?
This is an 11th grade problem, and I really don't know -after a lot of searching-what tools do I need to solve this problem using only high school knowledge.
|
Let $S_n$ denotes the random variable corresponding to the sum modulo $4$ after $n$ throws. You can model $S_n$ as a Markov Chain with $4$ states $\{0,1,2,3\}$ and initial value $S_0 = 0$
The matrix associated with this chain is :
$$M = \frac{1}{6}\begin{pmatrix}
1 & 2 & 2 & 1 \\
1 & 1 & 2 & 2 \\
2 & 1& 1 & 2 \\
2 & 2 & 1 & 1
\end{pmatrix}$$ and the desired probability is $m^{(n)}_{1,1}$ where $m^{(n)}_{i,j}$ are the coefficients of $M^n$
Let's denote $$M' = \begin{pmatrix}
1 & 2 & 2 & 1 \\
1 & 1 & 2 & 2 \\
2 & 1& 1 & 2 \\
2 & 2 & 1 & 1
\end{pmatrix}$$
It is a circulant matrix and it implies that
*
*$M'$ is a diagonalizable matrix
*The 4 eigenvalues $\lambda_i$ are given by the $P(\omega)$ where $\omega \in \mathbb{U}_4$ and $P(x) = \sum_{j=1}^4 m_{1,j}x^{j-1} = 1 + 2x + 2x^2 + x^3$
We obtain $\lambda_1 = 0$, $\lambda_2 = 6$, $\lambda_2 = -1-i$, $\lambda_3 = -1+i$
and $M'$ can be written $$M' = Pdiag(\lambda_1,\lambda_2,\lambda_3,\lambda_4)P^{-1}$$ and therefore
$$M'^n = Pdiag(\lambda_1^n,\lambda_2^n,\lambda_3^n,\lambda_4^n)P^{-1}$$
This shows that $m'^{(n)}_{1,1}$ is a linear combination of $\lambda_i^n$.
More precisely, it exists 4 constant $a,b,c,d$ such as :
$m'^{(n)}_{1,1} = a \times 0^n + b \times 6^n + c(-1+i)^n + d(-1-i)^n$
with $m'^{(n)}_{1,1} = 1,1,9,55$ for $n=0,1,2,3$ you can solve a $4 \times 4$ system and get $$m'^{(n)}_{1,1} = \frac14(0^n + 6^n + (-1-i)^n + (-1+i)^n)$$
And you can retrieve $$m^{(n)}_{1,1} = \frac1{4 \times 6^n}(0^n + 6^n + (-1-i)^n + (-1+i)^n)$$
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|
Without Calculator find $\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$
Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.
My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{aligned}
& \cos \left(50^{\circ}\right)<\cos \left(45^{\circ}\right) \\
\Rightarrow \quad & 2 \cos \left(50^{\circ}\right)<\sqrt{2} \\
\Rightarrow \quad & x<\sqrt{3}+\sqrt{2}<3.14
\end{aligned}$$
Now I am struggling hard to prove that $x>3$
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A calculus-based solution:
As you noticed, we only need to prove that $ \cos 50^{\circ}>\frac{3-\sqrt 3}{2}$.
Because of the identity $\cos 3x=4\cos^3x -3\cos x $, we conclude that $\cos 50^{\circ}$ is a root of the equation:
$$\cos 150 ^{\circ}=\frac{-\sqrt 3}{2}=4t^3-3t.$$
However, it is very easy to see that $4t^3-3t+\frac{\sqrt 3}{2}=0$ has three roots; one in the range $(-\infty, -\frac{1}{2} ]$, one in the range $[-\frac{1}{2}, \frac{1}{2} ]$, and the other in the range $[\frac{1}{2}, \infty)$.
Since $\cos 50^{\circ} >\cos 60^{\circ}=\frac{1}{2}$, we must have $\cos 50^{\circ} \in [\frac{1}{2}, \infty) $. Moreover $f(t)=4t^3-3t+\frac{\sqrt 3}{2}$ is increasing for $t>\frac{1}{2}$.
Therefore, we only need to show that $f(\frac{3-\sqrt 3}{2})<0=f(\cos 50^{\circ})$:
$$f(\frac{3-\sqrt 3}{2})=\frac{(3-\sqrt 3)^3}{2}-\frac{3(3-\sqrt 3)}{2}+\frac{\sqrt 3}{2} \\= \frac{((3-\sqrt 3)^2-3)(3-\sqrt 3)}{2}+\frac{\sqrt 3}{2} \\= \frac{(9-6\sqrt 3)(3-\sqrt 3)+\sqrt 3}{2}=\frac{45-26\sqrt 3}{2}<0. $$
Now, we are done because: $45^2=26^2 \times 3-3$.
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|
Showing that the power series converges at the endpoints of its interval of convergence The problem is the following:
Find the power series for $\frac{1}{2}\arctan^2(x)$ and show it converges in the endpoints of the interval of convergence.
My difficulty is in the convergence part.
The power series that I got:
$\displaystyle \frac{1}{2}\arctan^2(x) = \frac{x^2}{2} - \frac{x^4}{4}\left(\frac{1}{3} + 1\right) + \frac{x^6}{6}\left(\frac{1}{5} + \frac{1}{3} + 1\right) + ...$
The radius of convergence is $1$ $\implies$ $x\in(-1,1)$ it converges, but what about $x = 1$ or $x = -1$?
Since we have even powers, it is enough to show that it converges for $x = 1$.
The exercise gives a hint that says to use that:
$\displaystyle \left(\frac{1}{2n-1}+...+\frac{1}{5} + \frac{1}{3} + 1\right)<1+\frac{1}{2}\ln(2n-1)$
I was trying to find something convergent to compare using the innequality in the sum, but I didn't find anything.
Any help is appreciated.
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Let $a_n=\frac1{2n+2}\sum_{k=0}^n\frac1{2k+1}.$
You will easily check that $(a_n)$ is decreasing, and the hint serves to prove it tends to $0.$
Then apply the alternating series test.
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|
Inequality $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5} $ I have trouble with solving this inequality:
Prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5}$ for a,b,c>0.
Using Cauchy-Schwartz I got this: $\frac{a^3}{3b^3+c^3} +\frac{b^3}{3c^3+2a^3} +\frac{c^3}{3a^3+2b^3} \geq \frac{3}{5} $
Now I have to prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{a^3}{3b^3+c^3} +\frac{b^3}{3c^3+2a^3} +\frac{c^3}{3a^3+2b^3}$
Any help is appreciated!
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By Am-Gm inequality we have $$3ab^2\leq a^3+b^3+b^3$$ so $$3ab^2 +2c^3\leq a^3+2b^3+2c^3$$ so $${a^3 \over 3ab^2+2c^3} \geq {a^3\over a^3+2b^3+2c^3}$$ Let $x=a^3$, $y=b^3$ and $z=c^3$. We can assume that $x+y+z=1$, and so we have to prove $$f(x) +f(y) +f(z)\geq 3/5$$ where $$f(x) ={x\over 2-x}$$ The tangent at $x=1/3$ is $$t(x)={18\over 25}x- {1\over 25}$$ Since $f(x) \geq t(x)$ for all $x\in[0,1)$ we have $$f(x) +f(y) +f(z)\geq t(x) +t(y) +t(z) = 3/5$$
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|
When is it valid to convert "$\le$" or "$\ge$" to "$=$"? I don't understand why the conversion of ≤ to = in this proof in Spivak's Calculus is legitimate. (The conversion in the previous line from = to ≤ makes sense to me.) Could someone please explain or elaborate on how $(2)$ follows from $(1)$ below?
this proof is motivated by the observation that
$$
|a|=\sqrt{a^2} \text {. }
$$
We may now observe that
$$
\begin{align*}
(|a+b|)^2=(a+b)^2 & =a^2+2 a b+b^2 \\
& \leq a^2+2|a| \cdot|b|+b^2 \tag{1} \\
& =|a|^2+2|a| \cdot|b|+|b|^2 \tag{2} \\
& =(|a|+|b|)^2
\end{align*}
$$
|
In case the existing answers are still not clear...
*
*The presentation $$
\begin{align*}
{}& (|a+b|)^2\\
={}& (a+b)^2 \\
={}& a^2+2 a b+b^2 \\
\leq {}& a^2+2|a| \cdot|b|+b^2 \tag{1} \\
={}& |a|^2+2|a| \cdot|b|+|b|^2 \tag{2} \\
={}& (|a|+|b|)^2
\end{align*}
$$ is asserting the conjunction of these 5 statements:
\begin{align*}
(|a+b|)^2 &= (a+b)^2 \\
(a+b)^2 &= a^2+2 a b+b^2 \\
a^2+2 a b+b^2 &\leq a^2+2|a| \cdot|b|+b^2 \tag{1'}\\
a^2+2|a| \cdot|b|+b^2 &= |a|^2+2|a| \cdot|b|+|b|^2 \tag{2'}\\
|a|^2+2|a| \cdot|b|+|b|^2 &= (|a|+|b|)^2.
\end{align*}
*It is then natural to conclude (using transivity of = together with substitution of each side of the inequality) that $$(|a+b|)^2 \le (|a|+|b|)^2.\tag C$$
Note that $(C)$ is generally not explicitly asserted, but just tacitly understood by the reader.
*
*
*Lines $(1)$ and $(2)$ are not statements, and we are not asserting that $(1)$ implies $(2)$ (nor asserting that they are equivalent, for that matter):
\begin{align}
\leq {}& a^2+2|a| \cdot|b|+b^2 \tag{1} \\
={}& |a|^2+2|a| \cdot|b|+|b|^2. \tag{2}
\end{align}
*Neither is fragment $(2)$ replacing fragment $(1);$ the ≤ symbol is not being dropped and subsequently ignored!
*On the other hand, given \begin{align}
a<b\\
b=c\\
c>d,
\end{align} it is invalid to draw any of the conclusions
*
*$a<d$
*$a=d$
*$a>d;$
in the absence of more data, any of these 3 statements might be true.
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|
Solve over natural numbers: $m^3=2n^3+6n^2$. Functional equation gives rise to a diophantine equation! My question is basically to find all natural numbers $(m,n)$ such that
$m^3=2n^3+6n^2$
First for some background (this is not really that relevant but anyways):
I was trying to solve an olympiad functional equation
$f:\mathbb{N} \to \mathbb{N}$ such that $f(x+y)=f(x)+f(y)+3(x+y)\sqrt[3]{f(x)f(y)}$
Then we will get that $f(x)$ will always be a perfect cube so setting $f(x)=g(x)^3$ and then $P(1,1)$ gives $g(2)^3=2g(1)^3+6g(1)^2$. And then I noticed that if $g(1)=1$ we could just induct up to get $g(x)=x$ always, so this is where the question comes from, because if i prove the only solution in natural numbers to the diophantine in $(1,2)$ I will be done.
Now back to the diophantine, obvioulsy $(m,n)=(2,1)$ is a solution and wolfram alhpha says it is the only solution. For the past one hour or so I have been trying to prove this but have done basically nothing...
I have just been doing random subsitutions and making cases and stuff, like $m=2y$ so then we get $y \equiv x \mod 3$
and then making cases like if $x$ is divisible by $3$ or not but I have not even been able to rule out one case...
There are infact more integer solutions so I believe we could use some inequalities somehow but I am not sure how whatsoever...
Any help on this problem will be appreciated...
Thanks!
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This doesn't answer the diophantine equation $ m^3 = 2n^3 + 6n^2$ (which is hard), but answers the functional equation $f(x+y) = f(x) + f(y) + 3(x+y) \sqrt[3]{f(x) f(y) } $ when defined over all integers $ f: \mathbb{Z} \rightarrow \mathbb{Z}$.
With $ x = y = 0$, $f(0) = 0$.
Now, suppose there is some $ a \neq 0 $ such that $f(a) = 0$, then with $ y = a$, we get that $f(x+a) = f(x)$ , so we have a periodic function. If there is some $x$ with $f(x) \neq 0$, then $f(x+x+a) = f(x+x)$ gives us $ 3(x+x+a)\sqrt[3]{f(x)f(x) } = 3(x+x)\sqrt[3]{f(x) f(y)}$, which is a contradiction. Hence, the only solution is $f(x) = 0 \, \forall x$ (which doesn't give us a solution on the positive integers).
Otherwise, $f(x) \neq 0 \Leftrightarrow x \neq 0$.
We will show that $\forall x, y \neq 0$,
$$\frac{f(x) } { x^3 } = \frac{f(y) } { y^3}.$$
This is obvious if $ x = - y$ by definition.
Otherwise, let $ z = - x - y \neq 0$. So $-f(x) = f(y+z) = f(y) + f(z) - 3x \sqrt[3]{f(y) f(z) }$, or that $ f(x) + f(y) + f(z) = 3x \sqrt[3]{ f(y) f(z) } $.
Swapping $x$ and $y$, we get that $ f(y) + f(x) + f(z) = 3y \sqrt[3]{f(x) f(z) }$.
Hence, $ \frac{f(x) } { x^3 } = \frac{f(y) } { y^3}$ as claimed. This has value $f(1)$.
Finally, setting $f(x) = f(1)x^3$ with $f(1) \neq 0$ in the functional equation, we get
$$f(1) ( x+y)^3 = f(1) x^3 + f(1) y^3 + 3(x+y) f(1)^{2/3} xy$$
or that $3xy(x+y) ( f(1) - f(1) ^ { 2/3} ) = 0 $.
This gives us $ f(1) = 1$ so $ f(x) = x^3 \, \forall x \in \mathbb{N}$.
Note
*
*Instead of solving $ m^3 = 2n^3 + 6n^2$, we're solving $ 8n^3 = 2n^3 + 6 n^2$ which is much easier.
*We might be able to work around extending the function to the negative integers (EG replacing $f(z) = f(-x-y) = - f(x+y)$), but that solution presentation felt the most natural to me.
*In fact, this solution also works if the domain and range are the reals. We didn't use any facts about integers / number theory.
Ignore this:
Suppose that the function is only defined for the positive integers.
We extend the function to the non-negative integers by defining $f(0) = 0$.
It remains to check that for $ x, y \geq 0$, we still have $ f(x+y) = f(x) + f(y) + 3(x+y) \sqrt[3]{f(x)f(y)}$, which is obvious in the cases of $ x = 0 , y = 0 , x = y = 0$.
Again, we extend the function to the negative integers by defining $f(-x) = - f(x)$. We have to check that the functional equation holds, which is left as an exercise to the reader.
Of course, we now have a superset of solutions, and have to check which allow for solutions to the positive integers.
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Evaluate $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Evaluate $$\large{\int} \small{\sqrt{\frac{1+x^2}{x^2-x^4}} \space {\large{dx}}}$$
Note that this is a Q&A post and if you have another way of solving this problem, please do present your solution.
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Recall the half-angle identity
$$\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}. \tag{1}$$ This suggests the substitution $$x^2 = \cos 2\theta, \quad x \, dx = -\sin 2\theta \, d\theta$$ which yields for $0 < x < 1$
$$\begin{align}
\int \sqrt{\frac{1+x^2}{x^2-x^4}} \, dx
&= \int \frac{1}{x^2} \sqrt{\frac{1+x^2}{1-x^2}} \cdot x \, dx \\
&= \int \sec 2\theta \cot \theta \,(-\sin 2\theta) \, d\theta \\
&= - \int \tan 2\theta \cot \theta \, d\theta \\
&= - \int (1 + \sec 2\theta) \, d\theta \\
&= - \theta - \frac{1}{2} \log (\sec 2\theta + \tan 2\theta) + C \\
&= - \frac{1}{2} \arccos x^2 - \frac{1}{2} \log \left( \frac{1 + \sqrt{1-x^4}}{x^2} \right) + C.
\end{align} $$
|
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|
Find integral of $\int\frac{x^2+3}{\sqrt{x^2+6}}dx$ from Cambridge IGCSE Additional Mathematics $$\int\frac{x^2+3}{\sqrt{x^2+6}}dx$$
I tried to $$\int\frac{x^2+6-3}{\sqrt{x^2+6}}dx=\int\frac{x^2+6}{\sqrt{x^2+6}}dx-\int\frac{3}{\sqrt{x^2+6}}dx= \int\sqrt{x^2+6}dx - 3\int\frac{1}{\sqrt{x^2+6}}dx$$
Then got stuck
In exam-mate website the answer is $$\frac{1}{2}x\sqrt{x^{2}+6}$$
Thanks for replies in advance
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You can enforce an Euler substitution,
$$t=\frac{\sqrt{x^2+6}-\sqrt6}x \implies x=\frac{2\sqrt6\,t}{1-t^2} \implies dx = 2\sqrt6\frac{1+t^2}{(1-t^2)^2} \, dt$$
to transform the starting integral to
$$\int \frac{x^2+3}{\sqrt{x^2+6}}\,dx = 2\sqrt6 \int \frac{\frac{24t^2}{(1-t^2)^2}+3}{\frac{2\sqrt6\,t^2}{1-t^2}+\sqrt6} \, \frac{1+t^2}{(1-t^2)^2}\,dt = 6 \int \frac{1+6t^2+t^4}{(1-t^2)^3}\,dt$$
|
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|
Valid proof for integral of $1/(x^2+a^2)$ I'm trying to prove some integral table formulae and had a concern over my proof of the following formula:
$$\int\frac{1}{x^2+a^2}\;dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C$$
Claim:
$$\frac{1}{x^2+a^2}=\frac{1}{a^2}\sum_{k=1}^\infty(-1)^{k-1}\left(\frac{x}{a}\right)^{2k}\hspace{5mm}\forall\;x\in(-a,a)$$
Proof:
$$\frac{1}{x^2+a^2}=\frac{1}{a^2}-\frac{x^2}{a^4}+\frac{x^4}{a^6}-\frac{x^6}{a^8}+\dots$$
$$\implies 1=(x^2+a^2)\left(\frac{1}{a^2}-\frac{x^2}{a^4}+\frac{x^4}{a^6}-\frac{x^6}{a^8}+\dots\right)$$
$$\implies 1=\left(\frac{x^2}{a^2}-\frac{x^4}{a^4}+\frac{x^6}{a^6}-\frac{x^8}{a^8}+\dots\right)+\left(1-\frac{x^2}{a^2}+\frac{x^4}{a^4}-\frac{x^6}{a^6}+\dots\right)$$
$$\implies 1=1\hspace{5mm}\forall\;x\in(-a,a)$$
My proof then uses the following:
$$\int\frac{1}{x^2+a^2}\;dx=\frac{1}{a^2}\int\sum_{k=1}^\infty(-1)^{k-1}\left(\frac{x}{a}\right)^{2k}\;dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C$$
My concern is that this isn't a valid proof since the radius of convergence of arctangent's Taylor Series is finite. I'm 7 years removed from taking calculus so I'm admittedly forgetful of the fine details on this. Could someone explain if this is a valid approach or, if not, why?
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This is what I call an almost immediate antiderivative, based on the fact that if $\;\int f(x)\,dx = F(x)\;$ , then we can conclude that $\;\int f(g(x)) g'(x)\,dx = F(g(x))\;$, which of course it's just the basis of substitution, and thus:
$$\int\frac{dx}{a^2+x^2}=\frac1a\int\frac{\frac1adx}{1+\left(\frac xa\right)^2}=\frac1a\int\frac{d\left(\frac xa\right)}{1+\left(\frac xa\right)^2}=\frac1a\arctan\frac xa+C$$
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series
$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$
Does it converge? If so, what is its sum?
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There are actually two "more direct" proofs of the fact that this limit is $\ln (2)$.
First Proof Using the well knows (typical induction problem) equality:
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \,.$$
The right side is $\frac{1}{n} \left[ \frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+..+\frac{1}{1+\frac{n}{n}} \right]$ which is the standard Riemann sum associated to $\int_0^1 \frac{1}{1+x} dx \,.$
Second Proof Using $\lim_n \frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}-\ln (n) =\gamma$.
Then
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=
\left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n} \right]-2 \left[\frac{1}{2}+\frac{1}{4}...+\frac{1}{2n} \right] $$
$$= \left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n} \right]-\ln(2n) - \left[\frac{1}{1}+\frac{1}{2}...+\frac{1}{n} \right]+\ln(n) + \ln 2 \,.$$
Taking the limit we get $\gamma-\gamma+\ln(2)$.
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|
What is $\sqrt{i}$? If $i=\sqrt{-1}$, is $\large\sqrt{i}$ imaginary?
Is it used or considered often in mathematics? How is it notated?
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$$\sqrt{i}=\left|\sqrt{i}\right|e^{\arg\left(\sqrt{i}\right)i}$$
First we look to $\left|\sqrt{i}\right|$:
$$\left|\sqrt{i}\right|=\left|\sqrt{\frac{1}{2}+i-\frac{1}{2}}\right|=\left|\sqrt{\frac{1+(0+2i)-1}{2}}\right|=\left|\sqrt{\frac{1+2(0+1i)+(0+1i)^2}{2}}\right|=$$
$$\left|\sqrt{\frac{(1+(0+1i))^2}{2}}\right|=\left|\frac{\sqrt{(1+(0+1i))^2}}{\sqrt{2}}\right|=\left|\frac{1+(0+1i)}{\sqrt{2}}\right|=$$
$$\left|\frac{(1+(0+1i))\sqrt{2}}{2}\right|=\frac{\left|(1+(0+1i))\sqrt{2}\right|}{|2|}=\frac{\sqrt{2}|1+(0+1i)|}{2}=$$
$$\frac{|1+1i|}{2}=\frac{\sqrt{2}\sqrt{1^2+1^2}}{2}=\frac{\sqrt{2}\sqrt{2}}{2}=\frac{2}{2}=1$$
Now the argument of $\sqrt{i}$:
It's positive so on the complex axces, so $\sqrt{\sqrt{-1}}$ gives us $1e^{\frac{1}{4}\pi i}$ so the argument of $\sqrt{i}$ is $\frac{1}{4}\pi$
$-------$
$$\sqrt{i}=\left|\sqrt{i}\right|e^{\arg\left(\sqrt{i}\right)i}=1e^{\frac{1}{4}\pi i}=1\left(\cos\left(\frac{1}{4}\pi\right)+\sin\left(\frac{1}{4}\pi\right)i\right)=$$
$$\cos\left(\frac{1}{4}\pi\right)+\sin\left(\frac{1}{4}\pi\right)i=$$
$$\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$$
So:
$$\sqrt{i}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$$
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|
Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ How does one sum the series $$ S = a -\frac{2}{3}a^{3} + \frac{2 \cdot 4}{3 \cdot 5} a^{5} - \frac{ 2 \cdot 4 \cdot 6}{ 3 \cdot 5 \cdot 7}a^{7} + \cdots $$
This was asked to me by a high school student, and I am embarrassed that I couldn't solve it. Can anyone give me a hint?!
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You can use the formula
$\displaystyle \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx} = \frac{2 \cdot 4 \cdot 6 \cdots 2k}{3\cdot 5 \cdots (2k+1)}$
This is called Wallis's product.
So we have $\displaystyle S(a) = \sum_{k=0}^{\infty} (-1)^k a^{2k+1} \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx}$
Interchanging the sum and the integral
$\displaystyle S(a) = \int_{0}^{\frac{\pi}{2}}{\sum_{k=0}^{\infty}{(-1)^{k}(a\sin x)^{2k+1}} dx}$
The sum inside the integral is a geometric series of the form
$\displaystyle x - x^3 + x^5 - \cdots = x(1 - x^2 + x^4 - \cdots) = \frac{x}{1+x^2}$
Hence,
$\displaystyle S(a) = \int_{0}^{\frac{\pi}{2}}{\frac{a\sin x}{1 + (a\sin x)^2}}dx$
Now substitute $\displaystyle t = a \cos x$
The integral becomes
$\displaystyle \int_{0}^{a}{\frac{1}{1+a^2 - t^2}}dt = \frac{1}{2\sqrt{a^2+1}}\ln \left(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a} \right)$
Now $\displaystyle \ln \left(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a}\right) = \ln \left(\frac{\left(\sqrt{a^2+1}+a \right)^2}{\left(\sqrt{a^2+1}-a \right)\left(\sqrt{a^2+1}+a \right)}\right) = 2\ln \left(\sqrt{a^2+1}+a \right)$
So
$\displaystyle S(a) = \frac{1}{\sqrt{a^2+1}}\ln \left(\sqrt{a^2+1}+a \right) = \frac{\sinh^{-1}(a)}{\sqrt{a^2+1}}$
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|
sangaku - a geometrical puzzle
Find the radius of the circles if the size of the larger square is 1x1.
Enjoy!
(read about the origin of sangaku)
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Let $r$ be the length the radius of the circles, and let $\theta$ be the measure of the (smaller) angle made at the corner of the big square.
The width of the square is equal to two radii and the projection of a double diameter (a quadruple-radius), so that
$(1)\hspace{1.0in}4r\cos\theta=1-2r$
Looking at the four right triangles, we see that the center circle's diameter is equal to the difference in the lengths of the legs; since the hypotenuse has length $1$, we have
$(2)\hspace{1.0in}2r = \cos\theta - \sin\theta$
From here, we simply need to eliminate $\theta$.
Multiplying (2) through by $4r$ and substituting in from (1) ...
$$8 r^2 = 4r\cos\theta - 4r \sin\theta = 1 - 2r - 4r \sin\theta$$
$$4r \sin\theta = 1 - 2r - 8 r^2$$
Therefore,
$$\begin{eqnarray}16r^2 &=& (4r \cos\theta)^2 + (4 r \sin\theta)^2 \\\
&=& ( 1 - 2r )^2 + ( 1 - 2r - 8 r^2 )^2 \\\
&=& 2 - 8 r - 8 r^2 + 32r^3 + 64 r^4 \end{eqnarray}$$
so that
$$0 = 32 r^4 + 16 r^3 - 12 r^2 - 4 r + 1 = (2r+1)(2r-1)(8 r^2 + 4 r - 1)$$
The roots of the polynomial are $\pm1/2$ and $(-1\pm\sqrt{3})/4$. We can eliminate three of them from consideration to conclude that $r = (-1+\sqrt{3})/4$.
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basic combinatorics question Each person from a group of 3 people can choose his dish from a menu of 5 options. Knowing that each person eats only 1 dish what are the number of different orders the waiter can ask the chef?
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(5+2 choose 3)=(7 choose 3)=35 not (5 choose 3). The 35 different orders orders are {{1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 1, 4}, {1, 1, 5}, {1, 2, 2}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 3}, {1, 3, 4}, {1, 3, 5}, {1, 4, 4}, {1, 4, 5}, {1, 5, 5}, {2, 2, 2}, {2, 2, 3}, {2, 2, 4}, {2, 2, 5}, {2, 3, 3}, {2, 3, 4}, {2, 3, 5}, {2, 4, 4}, {2, 4, 5}, {2, 5, 5}, {3, 3, 3}, {3, 3, 4}, {3, 3, 5}, {3, 4, 4}, {3, 4, 5}, {3, 5, 5}, {4, 4, 4}, {4, 4, 5}, {4, 5, 5}, {5, 5, 5}}. The notation and formula were from http://www.oeis.org/A000292
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Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
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Taking log of $\rm\ sin^2(x)\ =\ 1 - cos^2(x)\ = (1-cos(x))\ (1+cos(x))\ $ shows both answers identical
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|
Bug in mathematica: computing the sum of the ratios $(n-k+1)/(n-k+2)$ While experimenting with certain sums, I came to the following sum:
$$S_n = \sum_{k=0}^n \frac{n-k+1}{(n+1) (n+2) (n-k+2)}.$$
After rewriting the summand as
$$\frac{n-k+1}{(n+1) (n+2) (n-k+2)} = \frac{n^{\underline{k}}}{(n+2)^{\underline{k}}(n-k+2)^2}$$
and then feeding the sum to Mathematica, i obtained
$$S_n = \frac{7}{6(n+1)(n+2)}$$
But directly summing the first expression above does not yield this closed-form result, because this "result" is wrong! (see edits below) Where is the bug or what am i missing?
EDIT: There seems to be some bug / feature in Mathematica which is leading to the erroneous computation above; the correct answer is of course, as also observed in two answers below. If someone can explain the bug, then he / she should feel free to report it to Wolfram.
(To be specific, the answer above is using $7/6 = (1+2)-H_{1+2}$, instead of $n+2 - H_{n+2}$ as it should)
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Here is the same answer with a variation on the calculations
$\begin{eqnarray*}
S_{n} &=&\sum_{k=0}^{n}\dfrac{n-k+1}{(n+1)(n+2)(n-k+2)} \\
&=&\dfrac{1}{(n+1)(n+2)}\sum_{k=0}^{n}\dfrac{n-k+1}{n-k+2}\qquad (n+1)(n+2)%
\text{ is independent of }k \\
&=&\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}\frac{m-1}{m}\qquad \text{substitution }m=n-k+2 \\
&=&\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}\left( 1-\dfrac{1}{m}\right) \\
&=&\dfrac{1}{(n+1)(n+2)}\left(\sum_{m=2}^{n+2}1\right)-\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}%
\dfrac{1}{m} \\
&=&\dfrac{n+2-2+1}{(n+1)(n+2)}-\dfrac{1}{(n+1)(n+2)}\left( \sum_{i=1}^{n+2}%
\dfrac{1}{i}\right) +\dfrac{1}{(n+1)(n+2)} \\
&=&\dfrac{n+1}{(n+1)(n+2)}-\dfrac{H_{n+2}}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)} \\
&=&\dfrac{n+2-H_{n+2}}{(n+1)(n+2)},
\end{eqnarray*}$
where
$H_{n+2}=\displaystyle\sum_{i=1}^{n+2}\dfrac{1}{i}.$
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|
Finding the least positive root How to find the least positive root of the equation $\cos 3x + \sin 5x = 0$?
My approach so far is to represent $\sin 5x$ as $\cos \biggl(\frac{\pi}{2} - 5x\biggr)$ then the whole equation reduces to $$2\cos \biggl(\frac{\pi}{4} - x\biggr)\cdot \cos \biggl(\frac{\pi}{4} - 4x\biggr) = 0$$
From here we can write:
$$\biggl(\frac{\pi}{4} - x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$
$$\biggl(\frac{\pi}{4} - 4x\biggr) = n\pi + \frac{\pi}{2} , n \in \mathbb{Z}$$
Now there can be infinitely many solutions for this, what I am not getting how to compute the minimum among them? And what about if I am asked to find the maximum?
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EDIT: So apparently you want a full solution...
A product of two real numbers is $0$ precisely when a factor is $0$, so you must have that
$$ \begin{align}
& \cos\left(\tfrac{\pi}{4} - x\right) = 0 \qquad\;\; (1) \text{, or} \\\\
& \cos\left(\tfrac{\pi}{4} - 4x\right) = 0 \qquad (2).
\end{align} $$
Now we use that $\cos a = 0 \iff a \in \tfrac{\pi}{2} + \pi\mathbb{Z}$:
$$ \begin{align}
\cos \left(\tfrac{\pi}{4} - x\right) = 0
& \iff \tfrac{\pi}{4} - x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\
& \iff x \in \tfrac{3\pi}{4} + \pi \mathbb{Z},
\end{align} $$
the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{4}$.
$$ \begin{align}
\cos \left(\tfrac{\pi}{4} - 4x\right) = 0
& \iff \tfrac{\pi}{4} - 4x \in \tfrac{\pi}{2} + \pi\mathbb{Z} \\\\
& \iff 4x \in \tfrac{3\pi}{4} + \pi \mathbb{Z} \\\\
& \iff x \in \tfrac{3\pi}{16} + \tfrac{\pi}{4}\mathbb{Z},
\end{align} $$
the smallest positive $x$ satisfying this clearly being $\tfrac{3\pi}{16}$. The problem is now reduced to picking the smallest of two real numbers.
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Generating function for an Arithmetic mean I want to find out, how many ways I can produce an average 4.6 of 13 numbers, when I have only numbers {1,2,3,4,5}.
I thought, that it could be done by generating functions, but since my knowledge on them isn't very deep yet, I'm a little lost. Also I have a hard time dealing with the average not being integer.
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If you are adding 13 numbers and the average is $4.6$, this is equivalent to their sum being $4.6\times 13$. Unfortunately, that gives $59.8$, which is impossible to achieve if your original numbers were all integers. Which suggests (to me, at least) that you are in fact approximating the average, and that the actual average you want is $\frac{60}{13} = 4.615384615\ldots$
So, you want to know how many ways you can obtain $60$ as a sum of exactly $13$ integers, all taken from ${1,2,3,4,5}$. This is equivalent to counting the number of $5$-partitions of $60$ of length exactly $13$.
It can probably be done with generating functions, but here you can do it from first principles.
Note that the maximum sum you can possibly get with $13$ numbers taken from ${1,2,3,4,5}$ is $65$. So the partitions you want are in 1-to-1 correspondence with the partitions of $5$ other than the trivial one ($5=5$).
For instance, the partition $5=1+4$ gives the partition $60 = 11(5) + 4 + 1$ (subtract $1$ from one summand, $4$ from another summand). The partition $5=2+3$ yields the partition $60=11(5) + 3 + 2$. Etc. So you have:
\begin{align*}
5 &= 1+4 &\qquad 60 &= 11(5) + 4 + 1;\\
5 &= 2+3 &\qquad 60 &= 11(5) + 3 + 2;\\
5 &= 1+1+3 &\qquad 60 &= 10(5) + 4 + 4 + 2;\\
5 &= 1+2+2 &\qquad 60 &= 10(5) + 4 + 3 + 3;\\
5 &= 1+1+1+2 &\qquad 60&= 9(5) + 4 + 4 + 4 + 3;\\
5 &= 1+1+1+1+1 &\qquad 60&= 8(5) + 4 + 4 +4 + 4 + 4;
\end{align*}
where $60 = k(5) +\cdots$ means that you take the number $5$ $k$ times, and then the other numbers.
So there are exactly $6$ ways of doing it.
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How to know that $a^3+b^3 = (a+b)(a^2-ab+b^2)$ Is there a way of go from $a^3+b^3$ to $(a+b)(a^2-ab+b^2)$ other than know the property by heart?
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There is a 'trick' for odd powered exponents as follows;
recall the identity
$ (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}) = x^n - y^n $ (*)
Now, there is a real number, a, such that y = (-a) [really this real number is -y but it is not important for the purposes of searching for a possible factorization].
For n even:
$y^n = (-a)^n = a^n$
but for odd n:
$y^n = (-a)^n = -a^n$
If we substitute -a for y in (*) we obtain (we assume n is odd)
$ x^n - y^n = x^n - (-a^n) = x^n + a^n = (x+a)(x^{n-1}-x^{n-2}a+...-xa^{n-2}+a^{n-1}) $
where we alternate + and - within the last bracket due to the parity of n-i.
When n = 3
$(x-y)(x^2+xy+y^2) = x^3-y^3$
which means
$x^3+y^3 = (x+y)(x^2-xy+y^2)$
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|
Calculation of inverse of chi-square's expectation I don't know where to begin to calculate the expectation value of the random variable $1/V$, where $V$ is a random variable with chi-square distribution $\chi^2(\nu)$.
Could somebody help me?
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I try to help this question.
A random variable $X$ with inverse chi-square distribution has p.d.f
$$\frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big), x>0$$
Since it is a proper distribution, we have
$$\int_0^{\infty} x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx=1 \rightarrow 2^{\frac{v}{2}} \Gamma(\frac{v}{2})$$
Therefore, the expectation for inverse chi-square distribution is:
$$E(X) = \int_0^{\infty} x \cdot \frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})}x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx = \frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} \int_0^{\infty} x^{-\frac{v-2}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx$$
$$\frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} \cdot 2^{\frac{v-2}{2}}\cdot \Gamma\Big(\frac{v-2}{2}\Big) = \frac{1}{2 \cdot \frac{v-2}{2}} = \frac{1}{v-2}$$
$$E(X^2) = \int_0^{\infty} x^2 \cdot \frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})}x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx = \frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} \int_0^{\infty} x^{-\frac{v-4}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx$$
$$\frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} \cdot 2^{\frac{v-4}{2}}\cdot \Gamma\Big(\frac{v-4}{2}\Big) = \frac{1}{2^2 \cdot \frac{v-2}{2} \cdot \frac{v-4}{2}} = \frac{1}{(v-2)\cdot (v-4)}$$
Therefore, the variance of inverse chi-square distribution is:
$$V(X) = E(X^2) - [E(X)]^2 = \frac{1}{(v-2)\cdot (v-4)} - \Big(\frac{1}{v-2}\Big)^2 = \frac{}{}\frac{2}{(v-2)^2 (v-4)}$$
Hope this helps!
|
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How to calculate the total number of possible triangles where the perimeter is given, and all sides are integers? How would I go about calculating the total number of possible triangles where the perimeter is given, and all sides are integers?
Exempli gratia:
*
*$a + b + c = 15$
*$d + e + f = 16$
*$x + y + z = 100$
|
Let ${\cal S}_p$ be the set of ordered triples $(a,b,c) \in \mathbb{N}^3$ such that $c \ge b \ge a \ge 1$, $a+b+c=p$, and $a+b > c$ (the triangle inequality). For instance, ${\cal S}_3 = \{ (1,1,1) \}$ and ${\cal S}_4 = \{ \}$. Each triple in ${\cal S}_{p-2}$ gives rise to a triple in ${\cal S}_{p}$ under the transformation $(a,b,c)\rightarrow(a,b+1,c+1)$, and all triples in ${\cal S}_{p}$ are produced by this transformation except for those of the form $(a,a,c)$. These latter triples can be counted directly: there is one for each value of $a$ for which $3a \le p < 4a$, and so there are exactly $I_p = \lfloor{p/3}\rfloor - \lceil{(p+1)/4}\rceil + 1$ of them. So we have the recurrence $S_p = S_{p-2} + I_p$ for the values of $S_p = |{\cal S}_p|$, with initial conditions $S_0=S_1=0$. The addend has period $12$, and takes on values
$$
I_{12k+(0,1,2,...,11)} = k + (0,0,0,1,0,0,1,1,0,1,1,1).
$$
We then have $S_{12(k+1)} = S_{12k} + 6k + 3$ and $S_{12(k+1)+1} = S_{12k+1} + 6k + 5$, leading to quadratic solutions $S_{12k} = 3k^2$ and $S_{12k+1} = 3k^2 + 2k$. The full set of solutions is
$$
\begin{eqnarray}
S_{12k} &=& 3k^2 \\
S_{12k+1} &=& 3k^2 + 2k \\
S_{12k+2} &=& 3k^2 + k \\
S_{12k+3} &=& 3k^2 + 3k + 1\\
S_{12k+4} &=& 3k^2 + 2k \\
S_{12k+5} &=& 3k^2 + 4k + 1 \\
S_{12k+6} &=& 3k^2 + 3k + 1 \\
S_{12k+7} &=& 3k^2 + 5k + 2 \\
S_{12k+8} &=& 3k^2 + 4k + 1 \\
S_{12k+9} &=& 3k^2 + 6k + 3 \\
S_{12k+10} &=& 3k^2 + 5k + 2 \\
S_{12k+11} &=& 3k^2 + 7k + 4.
\end{eqnarray}
$$
Note that, as noted elsewhere, for all odd $p$ we have $S_{p} = S_{p+3}$.
|
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|
Second-Order Linear Differential Equation I have the following differential equation:
$$y''+y=\cos(t)\cos(2t)$$
Maybe something can be done to $\cos(t)\cos(2t)$ to make it easier to solve. Any ideas?
Thanks in advance.
|
I don't see an elegant solution (and I fail to see how Arturo's comment helps immediately), but it can be done by brute force using the method of variation of parameters.
The homogeneous differential equation $y''(t) + y(t) = 0$ has the two fundamental solutions $y_{1}(t) = \sin{(t)}$ and $y_{2}(t) = \cos(t)$.
We must solve
\[
\begin{pmatrix}
0 \\
\cos(t)\cos(2t)
\end{pmatrix} =
\begin{pmatrix}
y_{1} & y_{2} \\
y_{1}^{\prime} & y_{2}^{\prime}
\end{pmatrix}
\begin{pmatrix}
u_{1} \\
u_{2}
\end{pmatrix} =
\begin{pmatrix}
\sin{t} & \cos{t} \\
-\cos{t} & \sin{t}
\end{pmatrix}
\begin{pmatrix}
u_{1} \\
u_{2}
\end{pmatrix}
\]
for $u_{1}$ and $u_{2}$ and find $U_{1}$ and $U_{2}$ such that $U_{1}^{\prime} = u_{1}$ and $U_{2}^{\prime} = u_{2}$.
A particular solution will then be given by
\[y_{0}(t) = U_{1} y_{1} + U_{2}y_{2}\qquad\text{and}
\qquad y=y_{0} + c_{1} y_{1} + c_{2}y_{2}\]
will be the general solution.
Since the matrix $A(t) = \begin{pmatrix}
\sin{t} & \cos{t} \\\
-\cos{t} & \sin{t}
\end{pmatrix}$ is orthogonal, its inverse is its transpose, so
\[
\begin{pmatrix}
u_{1} \\
u_{2}
\end{pmatrix}=
\begin{pmatrix}
\sin{t} & -\cos{t} \\
\cos{t} & \sin{t}
\end{pmatrix}
\begin{pmatrix}
0 \\
\cos(t)\cos(2t)
\end{pmatrix} =
\begin{pmatrix}
\cos^2(t)\cos(2t) \\
\sin(t)\cos(t)\cos(2t)
\end{pmatrix}
\]
Note that $u_{2}(t) = \sin(t)\cos(t)\cos(2t) = \frac{1}{2}\sin{(2t)}\cos{(2t)} = \frac{1}{4}\sin{(4t)}$, so we can take
\[U_{2}(t) = -\frac{1}{16}\cos{(4t)}.\]
On the other hand, using Arturo's comment three times, we get
\begin{align*}
u_{1}(t) & = \cos^2(t)\cos(2t) = \cos(t)\frac{1}{2}(\cos(3t)+\cos(t)) \\
& = \frac{1}{2} \cos{(t)}\cos(3t) + \frac{1}{2} \cos(t)\cos(t) = \frac{1}{4}
(\cos{(4t)} + \cos{(2t)}) + \frac{1}{4}(\cos{(2t)} + 1) \\
& = \frac{1}{4}\cos{(4t)} + \frac{1}{2} \cos{(2t)} + \frac{1}{4}.
\end{align*}
Integrating this gives
\[U_{1}(t) = \frac{1}{16} \sin(4t) + \frac{1}{4}\sin{(2t)} + \frac{1}{4}t.\]
Now we can simplify the particular solution
\begin{align*}
y_{0} = U_{1} y_{1} + U_{2}y_{2} = \cdots = \frac{1}{4}t \sin(t) - \frac{1}{16} \cos(3t) + \frac{5}{16}\cos(t)
\end{align*}
using some further trig identities (I'm really growing tired of this, sorry).
You should at least check yourself that $y_{0}^{\prime\prime} + y_{0} = \frac{1}{2}(\cos(t) + \cos(3t)) = \cos(t)\cos{(2t)}$ by Arturo's comment again.
Therefore the general solution is
\[
y(t) = y_{0} + c_{1} y_{1} + c_{2} y_{2} = \frac{1}{4}t \sin(t) - \frac{1}{16} \cos(3t) + \frac{5}{16}\cos(t) + c_{1}\sin{(t)} + c_{2} \cos(t)
\]
for some real constants $c_{1}$ and $c_{2}$.
|
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|
Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
Prove $(2n+1) + (2n+3) + (2n+5) +
\cdots + (4n-1) = 3n^{2}$ for all
positive integers $n$.
So the provided solution avoids induction and makes use of the fact that $1 + 3 + 5 + \cdots + (2n-1) = n^{2}$ however I cannot understand the first step: $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) = (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$. Once that has been established I can follow the rest, but I was hoping someone could help me understand why $(1 + 3 + 5 + \cdots + (4n-1))$ which already looks like less than the LHS can be made equal to the LHS by subtracting a positive number.
Additionally, I wanted to prove the equality using induction, but had trouble with that as well. I think I am thrown off by the last term $4n-1$. Even the initial case of $n=1$ is not totally clear to me: does it hold because $2(1)+1 = 3(1)^{2}$ or is it because $4(1)-1 = 3(1)^{2}$?
Either way, my approach was to replace every $n$ on the LHS with $n+1$ which resulted in: $$(2(n+1)+1) + (2(n+1)+3)+ \cdots + (4(n+1)-1)$$ and I am not sure what the second to last term in the sequence would be... I tried simplifying anyway $$(2n+3) + (2n+5) + \cdots + (4n+3)$$ and I this point I thought I could subtract $(2n+1)$ from both sides of the induction assumption resulting in $(2n+3) + (2n+5) + \cdots ? = 3n^{2} - (2n+1) - (4n-1)$ substituting this yields: $$3n^{2} - 2n - 1 - 4n + 1 + 4n + 3 = 3n^{2} + 6n + 3 - 8n + 1 = 3(n+1)^{2} - 8n + 1$$ but I guess I don't want the $-8n + 1$... I also tried substituting $3n^{2} - (2n+1)$ without that last term being subtracted, but that did not work out either. If anyone can help me understand how to do this properly I would really appreciate it. Thanks!
|
The first item, $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1)$
$ = (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$
comes because you are just adding and subtracting the same set of terms on the RHS.
Your base case for the induction has just one term on the left side, which I would write $2*1+1=3*1^2$ For the induction, assume $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1)=3n^2$ then extend it to $n+1$: $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) + (4n+1)+(4n+3)-(2n+1)$
$=3n^2+4n+1+4n+3-(2n+1)=3n^2+6n+1=3(n+1)^2$
|
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|
How to solve $x^3 + 2x + 2 \equiv 0 \pmod{25}$? My attempt was:
$x^3 + 2x + 2 \equiv 0 \pmod{25}$
By inspection, we see that $x \equiv 1 \pmod{5}$. is a solution of $x^3 + 2x + 2 \equiv 0 \pmod{5}$. Let $x = 1 + 5k$, then we have:
$$(1 + 5k)^3 + 2(1 + 5k) + 2 \equiv 0 \pmod{25}$$
$$\Leftrightarrow 125k^3 + 75k^2 + 25k + 5 \equiv 0 \pmod{25}$$
$$\Leftrightarrow 5 \equiv 0 \pmod{25}$$
And I was stuck here :( because k was completely cancelled out, how can we find solution for this equation?
Thanks,
|
Hint by others, we use Hensel's lemma. By Hensel's lemma we solve $x^{3}+2x+2=0$ (mod 5). This has solution $1$ and $3$. But $1$ satisfies $f'(1)=0$, by Chan's previous work it should be cast aside. Hensel's lemma suggests solutions of the form $3+5k$ with $3(3+5k)^{2}+2\not=0$(mod 5). Hensel's lemma give $k$ to be:
$k(29+90 k+75 k^2)=k^{3}+2k+2$ (mod 5)
Solve this we should get $k=2$. I do not know if this is the best way to solve it.
|
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|
Factorization of sum of two square The difference is $a^2 - b^2 = (a - b).(a + b)$
But what about when I have $a^{25} + 1$ ? According to wolfram alpha, the alternate form is:
*
*$(a+1) (a^4 -a^3 + a^2 -a + 1)( a^{20} - a^{15} + a^{10} -a^5 +1)$
However, the square root of 25 is a rational number 5.
But If I had 50, where the square root of 50 is irrational ?
*
*$(a^2 + 1) (a^8 -a^6 +a^4 -a^2 +1) (a^{40} -a^{30} +a^{20} -a^{10} +1 )$
In fact, I'm just wondering and trying to find out patterns, I'm very curious about that, since I haven't found anything related, only the alternate forms generated by wolfram. My question is how and what mathematical algorithm they used to find $k$ forms for $a^n + 1$ ?
Thanks in advance.
|
Such factorizations are special cases of general formulas for factorizations of cylotomic polynomials, e.g. see wikipedia. These formulas follow by Möbius inversion.
Such polynomial factorizations come in handy for integer factorization. For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian factorizations of cyclotomic polynomials $\rm\;\Phi_n(x) = C_n(x)^2 - n\ x\ D_n(x)^2\;$. These play a role in factoring numbers of the form $\rm\; b^n \pm 1\:$, cf. the Cunningham Project. Below are some simple examples of such factorizations:
$$\begin{array}{rl}
x^4 + 2^2 \quad=& (x^2 + 2x + 2)\;(x^2 - 2x + 2) \\\\
\frac{x^6 + 3^2}{x^2 + 3} \quad=& (x^2 + 3x + 3)\;(x^2 - 3x + 3) \\\\
\frac{x^{10} - 5^5}{x^2 - 5} \quad=& (x^4 + 5x^3 + 15x^2 + 25x + 25)\;(x^4 - 5x^3 + 15x^2 - 25x + 25) \\\\
\frac{x^{12} + 6^6}{x^4 + 36} \quad=& (x^4 + 6x^3 + 18x^2 + 36x + 36)\;(x^4 - 6x^3 + 18x^2 - 36x + 36) \\\\
\end{array}$$
|
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|
Prove $\sin(\pi/2)=1$ using Taylor series
Prove $\sin(\pi/2)=1$ using the Taylor series definition of $\sin x$,
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
It seems rather messy to substitute in $\pi/2$ for $x$. So we have
$$\sin(\pi/2)=\sum_{n=0}^{\infty} \frac{(-1)^n(\pi/2)^{2n+1}}{(2n+1)!}.$$
I'm not too sure where to go from here. Any help would be appreciated! Thanks!
|
[ A sketch: ]
Let $f(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$. You can easily show that
$$\begin{eqnarray}
f(x + c) & = & f(c) + f'(c)x + f''(c)\frac{x^2}{2!} + \cdots \quad(1) \\
f''(x) & = & -f(x) \quad(2)\\
f'(x) & = & 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \quad(3)
\end{eqnarray}$$
Then it is straight forward to prove that for any $c$: $$f(x+c) = f'(c)f(x) + f(c)f'(x) \quad(4)$$
We can also prove that there is a $b$ such that $f'(b) = 0$. Then $(1),(4) \Rightarrow f(b) = 1$ and so $$f(x+b) = f'(x) \quad(5)$$
Going a little further, $(2),(5) \Rightarrow f(x+4b) = f(x)$ so $b = \frac{1}{4}\mathrm{period}$.
But $f^2(x) + f'^2(x) = 1$ (all terms of expansion except the first of $f'(x)$ cancel out) so the period of $f$ is $2\pi$, $$b = \frac{\pi}{2}$$
|
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|
Show that $x^2 - 3y^2 = n$ either has no solutions or infinitely many solutions I have a question that I have problem with in number theory about Diophantine,and Pell's equations. Any help is appreciated!
We suppose $n$ is a fixed non-zero integer, and suppose that $x^2_0 - 3 y^2_0 = n$, where $x_0$ and $y_0$ are bigger than or equal to zero. Let $x_1 = 2 x_0 + 3 y_0$ and $y_1 = x_0 + 2 y_0$. We need to show that we have $x^2_1 - 3 y^2_1 = n$, with $x_1>x_0$, and $y_1>y_0$. Also, we need to show then that given $n$, the equation $x^2 - 3 y^2 = n$ has either no solutions or infinitely many solutions.
Thank you very much!
|
There are an infinite number of solutions to $x^2-3y^2=1$, which you can generate from the base solution $2^2-3\cdot 1^2 = 1$ as follows:
Given $(x,y) \in \mathbb{Z}^2$ such that $x^2-3y^2 = 1$:
\begin{align*}
x^2-3y^2 & =(x-\sqrt{3}y)(x+\sqrt{3}y)\\
&= 1 \\
(x-\sqrt{3}y)^n(x+\sqrt{3}y)^n &= (X - \sqrt{3}Y)(X+\sqrt{3}Y)\\
& = 1^n = 1
\end{align*}
for $(X,Y) \in \mathbb{Z}^2$.
Numbers of the form $a+\sqrt{3}b$ are closed under multiplication, so if we find one solution to $x^2-3y^2=(x-\sqrt{3}y)(x+\sqrt{3}y) = n$ and infinitely many pairs $(a,b)$ with $a^2-3b^2=1$, we can then generate more solutions with:
$(x -\sqrt{3}y)(a - \sqrt{3}b)(x+\sqrt{3}y)(a+\sqrt{3}b) = (n)(1) $
|
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|
Where's the trick in this partial fractions question? Here is a partial fractions question from page 287, q22 of Schaum's Calculus 5e:
$$ \int{ \frac{x^6 + 7\, x^5 + 15\, x^4 + 23\, x^2 + 25\, x - 3}{{\left(x^2 + 1\right)}^2\, {\left(x^2 + x + 2\right)}^2} dx} $$
Their answer: (verified)
$$\ln\!\left(\frac{x^2 + 1}{x^2 + x + 2}\right) + \frac{1}{\left(x^2 + x + 2\right)} - \frac{3}{\left(x^2 + 1\right)}$$
The partial fractions decomposition: (verified)
$$ \frac{\left(6\, x + 16\right)}{{\left(x^2 + 1\right)}^2} + \frac{\left(26\, x - 32\right)}{\left(x^2 + 1\right)} - \frac{\left(26\, x - 7\right)}{\left(x^2 + x + 2\right)} + \frac{\left(6\, x + 47\right)}{{\left(x^2 + x + 2\right)}^2} $$
(When I say "verified" I mean I used a math algebra package to verify it on a computer)
Even when working from the computer-done partial fractions decomposition, I still don't get the same answer. What gives? Why is their answer so simple (and correct?) How did they get it? Where's the trick??
|
The term $\rm\:32\ x^3\:$ is mistakenly omitted in the numerator of the integrand. Once you add that in then the partial fraction is
$$\rm - \frac{2\:x+1}{(x^2+x+2)^2} - \frac{2\:x+1}{x^2+x+2} + \frac{6\:x}{(x^2+1)^2} + \frac{2\:x}{x^2+1} $$
from which the claimed result follows easily.
|
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Mathematical Induction (product of $n$ consecutive numbers) Assumption:
$$(n+1)(n+2) \cdots (2n) = (2^n)\cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)$$
Prove for $n+1$:
$$(n+2)(n+3) \cdots (2(n+1)) = (2^{n+1}) \cdot 1 \cdot 3 \cdot 5 \cdots (2(n+1)-1)$$
Using the assumption, I divide both sides by $(n+1)$ and substitute RHS into my $n+1$ equation, however it does not equate.
|
Assumption
$$C(n) := (n+1)(n+2) \cdots (2n) = (2^n)\cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)$$
Proof
Basis:
$$2=2^1$$
Inductive step:
$$C(n+1):= (n+2)(n+3) \cdots (2(n+1)) = (2^{n+1}) \cdot 1 \cdot 3 \cdot 5 \cdots (2(n+1)-1)$$
$$We\ have\ to\ prove: C(n) \Rightarrow C(n+1)$$
$$\vdots$$
$$Little\ substitution:$$
$$(n+2)(n+3) \cdots (2n)(2n+1)(2n+2) = (n+1)(n+2) \cdots (2n) \cdot 2 \cdot (2n+1)$$
$$That\ leaves\ us\ with:$$
$$(2n+2) = 2 \cdot (n+1)$$
|
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Explaining an algebra step in $ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4)$ I have encountered this step in my textbook and I do not understand it, could someone please list the intermediate steps?
$$ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4). $$
Thanks,
|
Take out the common factor $(n+1)^{2}$ and simplify. So you get $$ (n+1)^{2} \cdot \biggl[ \frac{n^{2}}{4} + (n+1)\Bigr] = (n+1)^{2} \cdot \biggl[ \frac{n^{2}+4n + 4}{4}\biggr]$$
|
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|
Are these fractions all equal? Are the following expressions all equal to one another?
(2ab+c)/y
(2ab)/y + c/y
(2a)/y * (2b/y) + c/y
$$\dfrac{2ab+c}{y}=\dfrac{2ab}{y}+\dfrac{c}{y}=\dfrac{2a}{y}×\dfrac{2b}{y}+\dfrac{c}{y}$$
|
By starting with the first, and applying the distributive law, we get the second: $$\frac{2ab+c}{y} = (2ab+c) \frac{1}{y} = 2ab \frac{1}{y} + c \frac{1}{y} = \frac{2ab}{y} + \frac{c}{y}$$
The last one is not equal to the other two, because
$$\frac{2a}{y} \cdot \frac{2b}{y} + \frac{c}{y}= \frac{2a\cdot 2b}{y\cdot y} + \frac{c}{y} = \frac{4ab}{y^2} + \frac{c}{y},$$
and $\frac{4ab}{y^2} + \frac{c}{y}$ is (usually) not equal to $\frac{2ab}{y} + \frac{c}{y}$. Notice that we used the fact that
$$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}.$$
So the first two are equal, but not the third.
|
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Why can ALL quadratic equations be solved by the quadratic formula? In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.
Why can I use $x = \dfrac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?
|
The proof of the quadratic formula takes advantage of completing the square.
$$ax^2 + bx + c = 0, \ a \neq 0, \ a, \ b, \ c \ \in \mathbb R$$
$$ax^2 + bx = -c$$
$$x^2 + \dfrac{b}{a}x = -\dfrac{c}{a}$$
$$x^2 + \dfrac{b}{a}x + \left(\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$$
$$\left(x+\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \dfrac{b^2}{4a^2}$$
$$\left(x+\dfrac{b}{2a}\right)^2 = -\dfrac{4ac}{4a^2} + \dfrac{b^2}{4a^2}$$
$$\left(x+\dfrac{b}{2a}\right)^2 = \dfrac{-4ac+b^2}{4a^2}$$
$$\left(x+\dfrac{b}{2a}\right)^2 = \dfrac{b^2-4ac}{4a^2}$$
$$x+\dfrac{b}{2a} = \pm \sqrt{\dfrac{b^2-4ac}{4a^2}}$$
$$x+\dfrac{b}{2a} = \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$
$$x=-\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$
$$\boxed{x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}}$$
This concludes the proof.
|
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Integration of powers of the $\sin x$ I have to evalute
$$\int_0^{\frac{\pi}{2}}(\sin x)^z\ dx.$$
I put this integral in Wolfram Alpha, and the result is
$$\frac{\sqrt{\pi}\Gamma\left(\frac{z+1}{2}\right)}{2\Gamma\left(\frac{z}{2}+1\right)},$$
but I don't know why. If $z$ is a positive integer, then one can do integration by parts, many times. Eventually, this yields
$$\int_0^{\frac{\pi}{2}}(\sin x)^{2z}\ dx=\frac{(2z-1)!!}{(2z)!!}\frac{\pi}{2},$$
where $(2n-1)!!=1\cdot 3\cdots (2n-1)$, and $(2n)!!=2\cdot 4\cdots 2n$.
I appreciate your help.
|
Just, following Theo's hint
$$
\begin{align*}
\int_{0}^{\frac{\pi}{2}}{(\sin\psi)^x}d\psi&= \int_{0}^{\frac{\pi}{2}}{(\sin\psi)^{2\cdot \frac{1}{2}(x+1)-1}(\cos\psi)^{2\cdot \frac{1}{2}-1}}d\psi\\
&=\frac{1}{2}B\left( \frac{x+1}{2},\frac{1}{2} \right)\\
&= \frac{1}{2}\cdot \frac{\Gamma\left(\frac{x+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left( \frac{x}{2}+1 \right)}\\
&=\frac{\sqrt{\pi}\Gamma\left(\frac{x+1}{2}\right)}{2\Gamma\left( \frac{x}{2}+1 \right)}.
\end{align*}$$
|
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|
Find $\cos(x+y)$ if $\sin(x)+\sin(y)= a$ and $\cos(x)+\cos(y)= b$ Find $\cos(x+y)$ if $\sin(x)+\sin(y)= a$ and $\cos(x)+\cos(y)= b$.
|
Square and add the two to get $$2 + 2 \cos(x-y) = a^2 + b^2$$
$$2 \cos(x-y) = a^2 + b^2 - 2 $$
Square and subtract the two to get $$\cos(2x) + 2 \cos(x+y) + \cos(2y) = b^2 - a^2$$
Now, $$\cos(2x) + \cos(2y) = 2 \cos(x+y) \cos(x-y) = \cos(x+y) (a^2+b^2-2)$$
Hence, we get
$$\cos(x+y) (a^2 + b^2) = b^2 - a^2$$
Hence,
$$\cos(x+y) = \frac{b^2-a^2}{b^2+a^2}$$
|
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|
Rules for Factorisation? Basically presented with this, simplify
\begin{aligned}
{\Bigl(\sqrt{x^2 + 2x + 1}\Big) + \Bigl(\sqrt{x^2 - 2x + 1}\Big)}
\end{aligned}
Possible factorisations into both
\begin{aligned}
{\Bigl({x + 1}\Big)^2}, {\Bigl({x - 1}\Big)^2}
\end{aligned}
\begin{aligned}
{\Bigl({1 + x}\Big)^2} , {\Bigl({1 - x}\Big)^2}
\end{aligned}
Hence when simplified, answer has two possibilities. One independent of x, and the other not.
( Simplified Answers: 2x, 2 )
Why is one independent and the other not? If such is equal to 2, why then when, say x=2, the answer does not simplify to 2?
|
We know $\sqrt{x^2} = |x|$, so
$$\sqrt{x^2 + 2x + 1} + \sqrt{x^2 - 2x + 1} = |x + 1| + |x-1|.$$
|
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|
Find the minimum value of $(\frac{x^n}{n} + \frac{1}{x})$ for $n \ge 4$
Find the minimum value of $(\frac{x^n}{n} + \frac{1}{x})$ for $n \ge 4$.
One possible approach could be by first writing $$ \left(\frac{x^n}{n} + \frac{1} {x}\right) = \left( \frac{x^n}{n} + \frac{1}{nx} + \frac{1}{nx}+ \frac{1}{nx} + \text{upto n terms}\right) $$ then using the property $\mathrm{AM} \ge \mathrm{GM}$, we would get $$\left(\frac{x^n}{n} + \frac{1}{nx} + \frac{1}{nx}+ \frac{1}{nx}+ \text{upto }n \text{ terms}\right) = \left( \frac{x^n}{n} + \frac{1}{x} \right) \ge \frac{n+1}{n}$$ But this does not hold good for negative $x$; I am inquisitive to know how to approach for that case?
|
For $x \lt 0$ there is no minimum. As $x$ gets very close to $0$, the $\frac{1}{x}$ term gets very large and negative, faster than the $\frac{x^n}{n}$ can get positive (assuming $n$ is even-if $n$ is odd this term is negative, too).
|
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|
Local minimum and maximum of the function Can anyone help me to solve the following question?
maximize and minimize the function $(10-x)(10-\sqrt{9^2-x^2})$ over $x\in[0,10]$
This is a high school question, so is there any simple trick help solve it? Thanks!
|
We give a short argument for both maximum and minimum.
To bring out the symmetry, let $y=\sqrt{81-x^2}$.
We study the behaviour of $(10-x)(10-y)$, that is, of
$$xy-10(x+y)+100.\qquad\qquad\text{(Expression $1$)}$$
We will find the maximum and minimum values of Expression $1$, given that
$x^2+y^2=81$ and $x \ge 0$, $y \ge 0$.
Let $w=x+y$ and $xy=v$.
We are interested in the behaviour of
$$v-10w+100. \qquad\qquad\text{(Expression $2$)}$$
But since $(x+y)^2-2xy=x^2+y^2$, we have $w^2-2v=81$.
So we are interested in the behaviour of
$$\frac{w^2-81}{2} -10w +100, \quad\text{that is, of}$$
$$\frac{1}{2}\left(w^2-20w+119\right). $$
Complete the square. We get
$$\frac{1}{2}\left((w-10)^2+19\right).\qquad\qquad\text{(Expression $3$)}$$
Now it's over. There is a local minimum at $w=10$.
The minimum value is $19/2$.
The maximum is reached where $w$, that is, $x+y$, reaches a maximum subject to $x^2+y^2=81$. Since $(x+y)^2+(x-y)^2=2(x^2+y^2)=162$, the maximum value of $(x+y)^2$, and hence of $x+y$, occurs where $x=y$.
We can if we wish find the values of $x$ at which the minimum is reached. We need to solve the system $x+y=10$, $x^2+y^2=81$. That gives $2xy=10^2-81=19$, so $(x-y)^2=81-19=62$, and therefore $x-y =\pm \sqrt{62}$, and now we can find $x$ and $y$.
Note on Symmetry: All the way through, we have preserved symmetry between $x$ and $y$. We introduced symmetry in the initial setup, and every step involved only symmetric functions of $x$ and $y$. Symmetry allowed the easy identification of $x+y$ as a key parameter.
The formal algebraic symmetry comes, in this case, from the underlying geometry. For the problem posed by the OP is fundamentally geometric. It has to do with the interaction between a circle and a rectangular hyperbola.
|
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|
Two problems on number theory I need some ideas (preferable some tricks) for solving these two problems:
Find the largest number $n$ such that $(2004!)!$ is divisible by
$((n!)!)!$
For which integer $n$ is $2^8 + 2^{11} + 2^n$ a perfect square?
For the second one the suggested solution is like this : $ 2^8 + 2^{11} + 2^n = ((2^4)^2 + 2\times2^4\times2^6 + (2^ \frac{n}{2})^2 ) \Rightarrow n=12$
But I can't understand the approach,any ideas?
|
The second one is true. Because $$(2^{4} + 2^{6})^{2} = 2^{8} + 2 \cdot 2^{4} \cdot 2^{6} + 2^{12}$$ so your $n=12$.
As far as I know, the main idea is to write $2^{8}+2^{11}+2^{n}$ as $(2^4)^{2} + 2 \cdot 2^{4} \cdot x + x^{2}$. Then you will have to manipulate what $x$ is and intuition say $x=2^{6}$.
|
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|
Hard simultaneous equation problem $5x^2y-4xy^2+3y^3-2(x+y)=0$, $xy(x^2+y^2)+2=(x+y)^2$ How to solve this system of equations:
$$\begin{align*}
5x^2y-4xy^2+3y^3-2(x+y) &=0 \\
xy(x^2+y^2)+2 &=(x+y)^2
\end{align*}
$$
|
Notice that for every solution $(x, y)$, $(-x, -y)$ is also a solution.
The second equation admits factorization:
$$
(-1 + x y) (-2 + x^2 + y^2) = 0
$$
Now solve for $y$ and substitute into the other equation, and solve for $x$. Positive solutions resulting from this are $x=y=1$ and $x = 2 y = 2 \sqrt{\frac{2}{5}}$.
|
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|
How to find inverse of a composite function? I am stuck with this question,
Let $A=B=C=\mathbb{R}$ and consider the functions $f\colon A\to B$ and $g\colon B\to C$ defined by $f(a)=2a+1$, $g(b)=b/3$. Verify Theorem 3(b): $(g\circ f)^{-1}=f^{-1}\circ g^{-1}.$
I have calculated $f^{-1}$, $g^{-1}$, and their composition, but how do I find the inverse of $(g\circ f)$?
Here is how I have done so far,
$$\begin{align*}
\text{Let}\qquad\qquad b &= f(a)\\
a&= f^{-1}(b)\\
&{ }\\
b&=f(a)\\
b&=2a+1\\
\frac{b-1}{2} &= a\\
a &= \frac{b-1}{2}
\end{align*}$$
But $a=f^{-1}(b)$,
$$f^{-1}(b) = \frac{b-1}{2}.$$
${}$
$$\begin{align*}
\text{Let}\qquad\qquad a&=g(b)\\
b&= g^{-1}(a)\\
a&= g(b)\\
a &= b/3\\
b &= 3a\\
g^{-1}(a) &= 3a\qquad(\text{because }b=g^{-1}(a)
\end{align*}$$
${}$
$$\begin{align*}
f^{-1}\circ g^{-1} &= ?\\
&= f^{-1}\Bigl( g^{-1}(a)\Bigr)\\
&= f^{-1}(3a)\\
f^{-1}\circ g^{-1} &= \frac{3a-1}{2}
\end{align*}$$
$$\begin{align*}
g\circ f&= g\bigl(f(a)\bigr)\\
&= g(2a+1)\\
g\circ f &= \frac{2a+1}{3}\\
(g\circ f)^{-1} &= ?\\
\text{Let}\qquad\qquad &b=g\circ f
\end{align*}$$
EDIT:
Thanks for the answers, I followed the suggestions and came up with the answer,
Now I have two questions,
*
*The answers do match but the arguments are different. Is that ok?
*Is $(g\circ f)$ same as $(g\circ f(a))$?
|
You find the inverse of $g\circ f$ by using the fact that $(g\circ f)^{-1} = f^{-1} \circ g^{-1}$.
In other words, what gets done last gets undone first.
$f$ multiplies by 2 and then adds 1.
$g$ divides by 3.
Dividing by 3 is done last, so it's undone first.
The inverse first multiplies by 3, then undoes $f$.
Later note: Per the comment, to verify that $(g\circ f)^{-1} = f^{-1} \circ g^{-1}$:
Instead of confusingly writing $a = g(b)$, write $c=g(b)$. Then $c=b/3$, so $b=3c$, so
$$g^{-1}(c) = 3c.$$ And $$f^{-1}(b) = \frac{b-1}{2}.$$
So
$$
b = 3c\qquad\text{and}\qquad a = \frac{b-1}{2}.
$$
Put $3c$ where $b$ is and get
$$
a=\frac{3c-1}{2}.
$$
You want to show that that's the same as what you'd get by finding $g(f(a))$ directly and then inverting.
So $c = g(f(a)) = \dfrac{f(a)}{3} = \dfrac{2a+1}{3}$.
So take $c = \dfrac{2a+1}{3}$ and solve it for $a$:
$$
\begin{align}
3c & = 2a+1 \\
3c - 1 & = 2a \\ \\
\frac{3c-1}{2} & = a.
\end{align}
$$
FINALLY, observe that you got the same thing both ways.
|
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|
On a property of the binomial coefficient Let $n$ be a positive integer and $p$ a prime, how can I prove that the highest power of $p$ that divides $\binom{2n}{n}$ is exactly the number of $k\geq1$ such that $\lfloor 2n/p^k\rfloor$ is odd?
|
Let us write out this binomial coefficient a bit more
$$\binom{2n}{n} = \frac{2n \cdot (2n-1) \cdot \ldots \cdot (n+1)}{n \cdot (n-1) \cdot \ldots \cdot 1}$$
Above we have all numbers between $n + 1$ and $2n$, while below we have all numbers from $1$ to $n$. Now suppose $\lfloor 2n/p\rfloor = 2m + 1$. Then there are exactly $m + 1$ numbers between $(n+1)$ and $2n$ that are divisible by $p$, and $m$ numbers between $1$ and $n$ divisible by $p$. So this gives that $p \ | \ \binom{2n}{n}$, and in fact this gives exactly one factor $p$. Alternatively, if $\lfloor 2n/p\rfloor = 2m$ then the factors above and below cancel out, giving no remaining factor $p$.
Similarly, suppose $\lfloor 2n/p^2\rfloor = 2m + 1$. Then we get an extra factor $p^2$ above, compared to below the bar. However, we already counted the single factors $p$ when we checked the parity of $\lfloor 2n/p\rfloor$, so we only get one extra factor $p$. You can continue this for higher $k$, until at some point $p^k > 2n$ and $\lfloor 2n/p^k\rfloor = 0$ for all $k$.
If you want more of an intuition for this, consider e.g. $\binom{132}{66}$ and $p = 5$. Then $\lfloor 132/5\rfloor = 26$, so both above and below we get $13$ numbers divisible by $5$. These are $70, 75, \ldots, 130$ above and $5, 10, \ldots, 65$ below. So these factors $p$ cancel. Since $\lfloor 132/25\rfloor = 5$ we get $3$ numbers ($75, 100, 125$) above and $2$ numbers ($25, 50$) below contributing another $p$ to the product. Finally $125$ contributes another factor $p$ to the number, so $\binom{132}{66} = 5^2 q$, with $5 \not| \ \ q$.
Edit: You could perhaps make the above a bit more rigorous and shorter by explaining exactly how many factors $p$ are in the product $2n \cdot (2n-1) \cdot \ldots \cdot (n+1)$, and how many factors $p$ there are in the product $n \cdot (n-1) \cdot \ldots \cdot 1$. The number of factors $p$ in the product $(2n)!$ is simply:
$$\lfloor \frac{2n}{p} \rfloor + \lfloor \frac{2n}{p^2} \rfloor + \ldots + \lfloor \frac{2n}{p^k} \rfloor + \ldots$$
The number of factors in $n!$ is exactly
$$\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \ldots + \lfloor \frac{n}{p^k} \rfloor + \ldots$$
So the number of factors in $\binom{2n}{n}$ is
$$\left(\lfloor \frac{2n}{p} \rfloor + \lfloor \frac{2n}{p^2} \rfloor + \ldots + \lfloor \frac{2n}{p^k} \rfloor + \ldots\right) - 2\left(\lfloor \frac{n}{p} \rfloor + \lfloor \frac{n}{p^2} \rfloor + \ldots + \lfloor \frac{n}{p^k} \rfloor + \ldots\right)$$
You can then complete the proof by noting that
$$\lfloor \frac{2n}{p^k} \rfloor - 2 \lfloor \frac{n}{p^k} \rfloor = \begin{cases} 1 & \text{if } \lfloor 2n/p^k \rfloor \text{ odd} \\ 0 & \text{if } \lfloor 2n/p^k \rfloor \text{ even} \end{cases}$$
|
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|
Ramanujan's identity involving cube roots I have found this pretty identity involving cube roots, first stated by Ramanujan:
$$
\sqrt{m(4m-8n)^{\frac13} + n(4m+n)^{\frac13}}
= (4m+n)^{\frac23} + (4(m-2n)(4m+n))^{\frac13}- (2(m-2n))^{\frac23}.
$$
So I ask if there is any method to create similar identities?
|
The identity seems incorrect; instead it should read $(\ast)$
$$
\color{Red}{3} \cdot \sqrt{m(4m-8n)^{\frac13} + n(4m+n)^{\frac13}}
= (4m+n)^{\frac23} + (4(m-2n)(4m+n))^{\frac13}- \color{Red}{(2(m-2n)^2)^{\frac13}}.
$$
I might be asking for trouble by being the heretic =), but I find this particular identity quite unremarkable.
The key is the substitution
$$
u^3 = 4m+n, \quad v^3 = m-2n,
$$
motivated by the repeated occurrences of $(4m+n)^{1/3}$ and $(m-2n)^{1/3}$. Then
$$
m = \frac{2u^3 + v^3}{9}, \quad n = \frac{u^3 - 4v^3}{9}.
$$
Plugging in these in $(\ast)$,
$$
3 \cdot \sqrt{\frac{4^{1/3}(2u^3v+v^4) + (u^4 - 4uv^3)}{9}} \mathop{\stackrel{\color{Blue}{(?)}}{=}}
u^2 + 4^{1/3} uv - 4^{1/3} v^2
$$
$$
\iff u^4 + 2 \cdot 4^{1/3} \cdot u^3v - 4uv^3 + 4^{1/3} v^4 \mathop{\stackrel{\color{Blue}{(?)}}{=}}
\Big(u^2 + 4^{1/3} uv - 4^{1/3} v^2 \Big)^2,
$$
which can be verified by squaring the right hand side.
|
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Solving $8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$ I came up with this equation during my homework :
$8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$
My algebra is weak and I can't seem to find a way to solve for x nicely
Could someone please show me a decent way of doing this?
Thanks alot, Jason
|
Generally the best way is to just plough through the algebra (and algebra gets quite a bit more advanced than this!) :
*
*$8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$
*$4=x(1-\sqrt{5})+(1-x)(1+\sqrt{5})$ (dividing through by $2$ simplifies a lot of the subsequent terms)
*$4=x-x\sqrt{5}+1-x+\sqrt{5}-x\sqrt{5}$ (multiply out all the terms)
*$4=x-x\sqrt{5}-x-x\sqrt{5}+1+\sqrt{5}$ (rearrange to get all the $x$ terms out front)
*$4=x(1-\sqrt{5}-1-\sqrt{5})+1+\sqrt{5}$ (collect the x terms)
*$4=x(-2\sqrt{5})+1+\sqrt{5}$ (simplify)
*$(4-(1+\sqrt{5}))=x(-2\sqrt{5})$ (move the constant term to the left)
*$3-\sqrt{5} = x(-2\sqrt{5})$ (simplify the left)
*$x=(3-\sqrt{5})/(-2\sqrt{5})$ (divide both sides by $-2\sqrt{5}$)
*$x=3/(-2\sqrt{5}) + {1\over2}$ (split out the terms)
*$\displaystyle{x=-{3\sqrt{5}\over 10} + {1\over2}}$ (multiply the numerator and denominator of the first part through by $\sqrt{5}$)
Of course, I strongly recommend plugging this $x$ in to confirm that it satisfies your initial equation!
|
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
|
[All: This answer came here when I merged duplicates. Please take that into account when voting, JL]
You should prove before
$$
\sum_{i=0}^n i = \frac{n(n+1)}{2}
$$
Cases $n=0,1$ are trivial. Suppose it's true for $n-1$ then
\begin{align}
\sum_{i=0}^n i^3 &= n^3 + \sum_{i=0}^{n-1}i^3 \\
&=n^3 + \left(\sum_{i=0}^{n-1} i\right)^2 \\
&= n^3 + \frac{n^2(n-1)^2}{4}\\
& = \frac{4n^3 + n^4 + n^2 + -2n^3}{4}\\
&= \frac{(n+1)^2n^2}{4} \\
&= \left(\sum_{i=0}^{n} i\right)^2
\end{align}
|
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|
For vectors in three-dimensional space, if $a \cdot b$ and $a \cdot c$ are equal, and $a \times b$ and $a \times c$ are equal, are b and c equal? For vectors in three-dimensional space, if $a \cdot b$ and $a \cdot c$ are equal, and $a \times b$ and $a \times c$ are equal, are b and c equal? I tried looking for counter-examples or using coordinate-by-coordinate proofs, but that didn't get me anywhere.
|
We can conclude that $\mathbf b= \mathbf c$ provided $\mathbf a \neq \mathbf 0$.
Write $\mathbf d = \mathbf b - \mathbf c$. The given equations can be written as
$$
\mathbf a \cdot \mathbf d = 0, \ \ \
\mathbf a \times \mathbf d = \mathbf 0.
$$
Now, suppose $\theta$ is the angle between $\mathbf a$ and $\mathbf d$.
$$
|\mathbf a \cdot \mathbf d| = |\mathbf a| |\mathbf d| |\cos \theta|, \ \ \
|\mathbf a \times \mathbf d| = |\mathbf a| |\mathbf d| |\sin \theta|, \ \ \
$$
Then, squaring and adding,
$$
0 = 0 + 0 = |\mathbf a \cdot \mathbf d|^2 + |\mathbf a \times \mathbf d|^2 = |\mathbf a|^2 |\mathbf d|^2 (\cos^2 \theta + \sin^2 \theta) = |\mathbf a|^2 |\mathbf d|^2,$$
which implies that either $\mathbf a = \mathbf 0$ or $\mathbf d = \mathbf 0$. Finally, $\mathbf d = \mathbf 0 \iff \mathbf b = \mathbf c$.
|
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|
Twin primes of form $2^n+3$ and $2^n+5$ How to prove that $2^n+3$ and $2^n+5$ are both prime for only finitely many integers $n$?
And how to prove that there are infinitely many primes of the form $2^n+3$ and $2^m+5$
|
Settling the twin prime part of the question seems to go as follows:
1) If $2\mid n$, then $3\mid 2^n+5$, so it suffices to study odd values of $n$.
2) If $n\equiv 1\pmod 4$, then $5\mid 2^n+3$, so it suffices to study the case $n\equiv 3\pmod 4$.
3) If $n\equiv 1\pmod 3$, then $7\mid 2^n+5$. If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$, so for there to be infinitely many twin primes of this form we must have $3\mid n$.
Combinining items 2 and 3 above, we see that it suffices to exclude the possibility of infinitely many twin primes of this form, when $n\equiv 3\pmod {12}$. But...
4) If $n\equiv 3\pmod {12}$, then $13\mid 2^n+5$.
Looks like $(5,7)$ and $(11,13)$ are all.
|
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Easy system in five equations An interesting little problem:
I have found solutions $[a=3,b=2,c=1,d=5,e=4]$ but not able to find proof that these are all that exist.
Find, with proof, all integers $a$, $b$, $c$, $d$ and $e$ such that:
$a^2 = a + b - 2c + 2d + e - 8$
$b^2 = -a - 2b - c + 2d + 2e - 6$
$c^2 = 3a + 2b + c + 2d + 2e - 31$
$d^2 = 2a + b + c + 2d + 2e - 2$
$e^2 = a + 2b + 3c + 2d + e - 8$
It seems simple, but I cannot find a proof.
nikolai
|
Calculate the sum $(a-3)^2+(b-2)^2+(c-1)^2+(d-5)^2+(e-4)^2$.
|
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|
divisibility of an expression by 30 let $g$ be a natural number how to show that $30$ divides $(-8 g^5+20 g^4-50 g^3+115 g^2-167 g+90)$?
my guess:
$30$ divides $90$ so it is enough to show that $30|-8 g^5+20 g^4-50 g^3+115 g^2-167 g = g(-8 g^4+20 g^3-50 g^2+115 g-167) $ now if $30|g$ we are done, otherwise we have to show that
$30|-8 g^4+20 g^3-50 g^2+115 g-167$ , I don't know how to go further..
|
Note that for any $n\in\mathbb{Z}$,
$$30\mid n\iff 2\mid n\text{ and }3\mid n\text{ and }5\mid n.$$
Then note that
$$-8g^5+20g^4-50g^3+115g^2-167g+90\equiv0+0+0+g^2+g+0\equiv g+g\equiv 0\bmod 2$$
because, by Fermat's Little Theorem, $g^2\equiv g\bmod 2$. Similarly,
$$-8g^5+20g^4-50g^3+115g^2-167g+90\equiv g^5+2g^4+g^3+g^2+g+0\equiv $$
$$g+2g^2+g+g^2+g+0\equiv 0\bmod 3$$
because $g^3\equiv g\bmod 3$, and
$$-8g^5+20g^4-50g^3+115g^2-167g+90\equiv 2g^5+0+0+0+3g+0\equiv $$
$$2g+0+0+0+3g+0\equiv 0\bmod 5$$
because $g^5\equiv g\bmod 5$.
Thus, we have shown that any number of the form $-8g^5+20g^4-50g^3+115g^2-167g+90$ is divisible by 2, 3, and 5, and therefore is divisible by 30.
|
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|
The limit of truncated sums of harmonic series, $\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}$ What is the sum of the 'second half' of the harmonic series?
$$\lim_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?$$
More precisely, what is the limit of the above sequence of partial sums?
|
For every $x\in [n,n+1]$, with $n\ge1$, we have $\displaystyle \frac{1}{n+1} \le \frac{1}{x} \le \frac{1}{n}$. If we integrate over $[n,n+1]$, we get
\begin{align}
\int_{n}^{n+1}\frac{1}{n+1}dx &\le& \int_{n}^{n+1}\frac{1}{x}dx &\le& \int_{n}^{n+1}\frac{1}{n}dx\\
\frac{x}{n+1}\Big|_{n}^{n+1} &\le& \ln(x)\Big|_{n}^{n+1}&\le& \frac{x}{n}\Big|_{n}^{n+1}\\
\frac{n+1-n}{n+1} &\le& \ln(n+1)-\ln(n)&\le& \frac{n+1-n}{n}\\
\end{align}
that is
$$
\frac{1}{n+1} \le \ln(n+1)-\ln(n) \le \frac{1}{n}
$$
Taking the sum from $k+1$ to $2k$, we get
$$
S_{k}+\frac{1}{2k+1}-\frac{1}{k+1} \le \ln\left(\frac{2k+1}{k+1}\right) \le S_{k},
$$
where
$$
S_{k}=\sum_{n=k+1}^{2k}\frac{1}{n}
$$
After rearranging, we get
$$
\ln\left(\frac{2k+1}{k+1}\right) \le S_{k} \le \ln\left(\frac{2k+1}{k+1}\right)+\frac{1}{k+1}-\frac{1}{2k+1}.
$$
Taking the limit, and using the Squeeze Theorem, we deduce that
$$
\lim_{k\to \infty}S_{k}=\ln(2).
$$
|
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|
Ring such that $x^4=x$ for all $x$ is commutative Let $R$ be a ring such that $x^4=x$ for every $x\in R$. Is this ring commutative?
|
Here's an old post of mine from Yahoo! Answers:
First, note $-x = (-x)^4 = x^4 = x$, so $x+x = 0$ for any $x$ in $R$. Then
$(x^2+x)^2 = x^2 + x + x^3 + x^3 = x^2+x$. Thus $x^2+x$ is idempotent, and it is easy to see idempotent elements are central in this ring. [I give a proof of this at the end.]
Now let $x=a+b$, where $a$ and $b$ are arbitrary. From above, for any $c$ in $R$,
$c(x^2 + x) = (x^2 + x)c$, and expanding this out and cancelling terms we get
$c(ab + ba) = (ab + ba)c$. Setting $c=a$, we get, after cancelling again,
$a^2b = ba^2$. Thus, for any $x$ in $R$, $x^2$ is central. Then of course
$x = (x^2+x)-x^2$ is central.
To prove that idempotents are central, first note that if $xy=0$, then
$yx = (yx)^4 = y (xy)(xy)(xy)x = 0$. So now if $z^2 = z$, then
$z(y - zy) = 0$, so $(y-zy)z = 0$, or $yz = zyz$. Similarly,
$(yz - y)z = 0$, so $z(yz-y) = 0$, or $zy = zyz$. Thus $yz = zy$.
|
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|
Find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$ I want to find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$.
My thoughts so far: I want to find $p$ such that $ \left( \frac{15}{p} \right) = 1$. By multiplicativity of the Legendre symbol, this is equivalent to $ \left( \frac{5}{p} \right) \left( \frac{3}{p} \right) = 1 $. Using the Law of Quadratic Reciprocity, this is equivalent to finding $p$ such that $ - \left( \frac{p}{5} \right) \left( \frac{p}{3} \right) = 1$.
So there are two cases:
*
*$ \left( \frac{p}{5} \right) = -1, \left( \frac{p}{3} \right) = 1$.
*$ \left( \frac{p}{5} \right) = 1, \left( \frac{p}{3} \right) = -1 $.
For case (1.), the quadratic residues modulo $5$ are $1$ and $4$, so for $ \left( \frac{p}{5} \right ) = -1$, we must have that $p$ is $2$ or $3$ modulo $5$. We must also have that $p$ is $1$ modulo $3$ from the other condition. One of these pairs is incompatible, and we can solve to give $p$ is $13$ modulo $15$.
Similarly for case (2.)
Is this the correct approach? I'm unsure if each step in my working is an "if and only if". If $p$ is $13$ modulo $15$, is $15$ necessarily a quadratic residue modulo $p$?
Thanks!
|
I think the law of quadratic reciprocity hasn't been fully applied. From Quadratic Reciprocity,
$$
\left(\frac{15}{p}\right)=\left(\frac{3}{p}\right)\left(\frac{5}{p}\right)=(-1)^{(p-1)/2}\left(\frac{p}{3}\right)\left(\frac{p}{5}\right).
$$
There are now two cases. If $p\equiv 1\pmod{4}$, you have $(15|p)=(p|3)(p|5)$, so you want $(p|3)$ and $(p|5)$ to have the same sign. Then the squares modulo $3$ and $5$ are $p\equiv 1\pmod{3}$ and $p\equiv 1,4\pmod{5}$, and the nonsquares are $p\equiv 2\pmod{3}$ and $p\equiv 2,3\pmod{5}$. Using the extra condition that $p\equiv 1\pmod{4}$, you can just check that the only possibilities are
$$
p\equiv 1,17,49,53\pmod{60}
$$
based on which conditions you choose.
The second case is $p\equiv 3\pmod{4}$, and you now want $(p|3)$ and $(p|5)$ to have different signs. Again writing out the squares and nonsquares modulo $3$ and $5$, you have $p\equiv 1\pmod{3}$ and $p\equiv 2,3\pmod{5}$, or $p\equiv 2\pmod{3}$ and $p\equiv 1,4\pmod{5}$. Now using the condition that $p\equiv 3\pmod{4}$, the only possibilities are
$$
p\equiv 7,11,43,59\pmod{60}.
$$
|
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|
Asymptotics of $1^n + 2^{n-1} + 3^{n-2} +\cdots + (n-1)^2 + n^1$ Suppose $n\in\mathbb{Z}$ and $n > 0$.
Let $$H_n = 1^n + 2^{n-1} + 3^{n-2} +\cdots + (n-1)^2 + n^1.$$
I would like to find a Big O bound for $H_n$. A Big $\Theta$ result would be even better.
|
My approach is very much along the same line as that of leonbloy.
Let $f(k) = k^{n+1-k}$. Use Euler-Maclaurin formula:
$$
\begin{eqnarray}
\sum_{k=1}^n f(k) &=& \int_1^n f(x) \mathrm{d} x + \frac{1}{2} \left( f(1) + f(n) \right) + \sum_{m=1}^q \frac{B_{2m}}{(2m)!} \left( f^{(2m-1)}(n) - f^{(2m-1)}(1) \right) \\
&&+ \frac{1}{(2q)!} \int_1^n B_{2q}(\{t\}) f^{(2q)}(t) \mathrm{d} t
\end{eqnarray}
$$
Since $f(1) = 1$ and $f(n)=n$. Asymptotically, $f^{(2m-1)}(1) \sim -n \cdot \log^{2m-1}(n)$ and $f^{(2m-1)}(n) \sim n^{2m-1}$.
Thus the main term comes from the integral. Let $x=1+(n-1)t$.
$$
\int_1^n x^{n+1-x} \mathrm{d} x = (n-1) \int_0^1 \left( 1 + (n-1) t \right)^{t+n(1-t)} \,\, \mathrm{d} t
$$
The integrand is sharp-peaked with peak location at
$$
t_0 = \frac{1}{n-1} \left( \frac{n+1}{W((n+1) \, \mathrm{e})} -1 \right)
$$
Logarithm of the integrand, expanded in the vicinity of $t_0$:
$$
\left( n + t - n t \right) \log\left(1 + (n-1) t \right) = (n+1) \left( w_n + \frac{1}{w_n} - 2 \right) - \frac{\sigma_n}{2} ( t-t_0)^2 + o((t-t_0)^2)
$$
where $w_n = W((n+1)\mathrm{e})$ and $\sigma_n = \frac{(n-1)^2 w_n ( w_n + 1)}{n+1}$.
Thus
$$
\sum_{k=1}^n k^{n+1-k} \sim \exp\left( (n+1) \left( w_n + \frac{1}{w_n} - 2 \right) \right) \sqrt{\frac{2\pi (n+1)}{w_n (1+w_n)}}
$$
Here is the numerical simulation, showing the agreement on logarithmic scale:
|
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|
Prove by induction: $2^n + 3^n -5^n$ is divisible by $3$ Let $P(n) =
2^n + 3^n - 5^n
$.
I want to prove that $P(n)$ is divisible by $3$ for all integers $n\geq 1$.
The basis step for this proof is easy enough: $P(1)$ is divisible by $3$.
For the inductive step, I let $k$ be an arbitrary integer, then assume $P(k)$ is divisible by $3$, and set out to prove that $P(k+1)$ is divisible by $3$.
$$
P(k) = 2^k + 3^k - 5^k
$$
$$\begin{align*}
P(k + 1) &= 2^{k+1} + 3^{k+1} - 5^{k+1}\\
&= 2*2^k + 3*3^k - 5*5^k
\end{align*}
$$
I'm guessing that the best way to do this is to prove $P(k+1) - P(k)$ is divisible by $3$, but I'm not sure on that so this could be where I start to approach this wrong.. I'm not sure what else to try though.. $P(k+1) * P(k)$? But that wouldn't distribute very well would it?
So what I did is write out P(k+1) - P(k):
$$
P(k+1) - P(k) = 2^k + 2*3^k - 6*5^k
$$
At this point, I know that the second and third terms are divisible by 3, but I know that $2^k$ is not necessarily divisible by 3, so here I am stuck...
|
As an alternative to Rankeya's answer, you have:
$$P(k+1) = 2^{k+1} + 3^{k+1} - 5^{k+1}.$$
Then, proceeding as you do, we have:
$$P(k+1) = 2\cdot 2^k + 3\cdot 3^k - 5\cdot 5^k.$$
At this point, you want to use your induction hypothesis. Notice that you have enough $2^k$s, $3^k$s and $5^k$s for two $2^k+3^k-5^k$, with some stuff left over. That is:
$$P(k+1) = 2(2^k + 3^k - 5^k) + 3^k - 3\cdot 5^k.$$
But $3^k$ is of course a multiple of $3$, and $3\cdot 5^k$ is a multiple of $3$. And $2(2^k+3^k-5^k)$ is a multiple of $3$ by the induction hypothesis. So $P(k+1)$ is a sum of multiples of $3$, hence is a multiple of $3$.
|
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|
The equation $(1+x)^2\frac{dy}{dx}-xy=x^2y^2$ I am very grateful for all your comments and answers to my previous question concerning ODEs ( The equation $(x-2xy-y^2)\frac{dy}{dx}+y^2=0$ ). Now I am struggling with this one
$$(1+x)^2\frac{dy}{dx}-xy=x^2y^2.$$
It seems not to be hard but nevertheless all the tricks I know fail in this case.
Best regards,
D.
|
Step 1.
Let $y(x) = \frac{1}{w(x)}$. Then $y^\prime(x) = -\frac{1}{w(x)^2}\cdot w^\prime(x)$, so the differential equation becomes $ \frac{(1+x)^2 w^\prime(x) + x w(x) + x^2}{w(x)^2} = 0$, that is $ w^\prime(x) + \frac{x}{(1+x)^2} w(x) + \frac{x^2}{(1+x)^2}=0$.
Step 2. Integrating factor, let $w(x) = f(x) g(x)$, and we will seek $f(x)$ so as to get rid of the term $x w(x)$. Substituting, $f^\prime(x) g(x) + f(x) g^\prime(x) + \frac{x}{(1+x)^2}f(x) g(x) + \frac{x^2}{(1+x)^2} = 0$. Choosing $f(x)$ such that $f^\prime(x) + \frac{x}{(1+x)^2} f(x) = 0$, the equation becomes $ g^\prime(x) = -\frac{1}{f(x)} \frac{x^2}{(1+x)^2}$, which is readily solvable for a known $f(x)$.
Step 3. Solve the auxiliary equation $f^\prime(x) = -\frac{x}{(1+x)^2} f(x)$. For this, rewrite it as $\left( \log( f(x) )\right)^\prime = - \frac{x}{(1+x)^2}$, and integrate both sides, giving $\log(f(x)) = \int \frac{x}{(1+x)^2} \mathrm{d} x + C = \log(1+x) - \frac{x}{1+x} + \log f_0$. That is $f(x) = f_0 (1+x) \exp\left( - \frac{x}{1+x} \right)$.
Step 4. Solve the ODE for $g(x)$ by direct integration:
$$\begin{eqnarray}
g(x) &=& -\int \frac{1}{f(x)} \frac{x^2}{(1+x)^2} \mathrm{d} x + C = - \int \frac{1}{f_0} \exp\left( \frac{x}{1+x} \right) \frac{x^2}{(1+x)^3} \mathrm{d} x + C \\
&\stackrel{x=-\frac{t+1}{t}}{=}& \frac{\mathrm{e}}{f_0} \int \mathrm{e}^{t} \frac{t^2+2 t + 1}{t} \mathrm{d} t + C = \frac{\mathrm{e}}{f_0} \mathrm{e}^t (1+t) + \frac{\mathrm{e}}{f_0} \int \frac{\mathrm{e}^t}{t} \mathrm{d} t + C
\end{eqnarray}
$$
The remaining integral is not elementary, let's denote it by $F(t) = \int \mathrm{e}^t \frac{\mathrm{d} t}{t}$.
Step 5. Find $y(x) = \frac{1}{w(x)} = \frac{1}{f(x) g(x)}$.
$$
y(x) = \frac{\mathrm{e}^{\frac{x}{1+x}}}{(1+x)\left( \exp\left(\frac{x}{1+x} \right) \frac{x}{1+x} + F\left(-\frac{1}{1+x}\right) + C\right)} = \frac{1}{x + (1+x) \exp\left( - \frac{x}{1+x} \right) \left( F\left( -\frac{1}{1+x} \right) + C \right) }
$$
Step 6. Check your equation.
Are you sure the original equation is correct ? Should you have $(1+x^2) y^\prime(x)$, instead of $(1+x)^2 y^\prime(x)$, the solution would come out elementary.
|
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|
probability of number of draws A box contains 8 tickets. Two are marked 1, two marked 2, two marked 3, and two marked 4.
Tickets are drawn at random from the box without replacement until a number appears that has appeared before. Let X be the number of draws that are made.
so the probability when X = 1 is: $4*\frac{2}{8}*\frac{1}{7}$ which is $\frac{1}{7}$.
when X = 2, in my understanding, the probability should be $4*\frac{2}{8}*\frac{6}{7}*\frac{1}{6}$, which is $\frac{1}{7}$ again. I doubt my answer since I don't think the two probabilities will be the same, but I couldn't spot where my reasoning went wrong. Any help will be appreciated. Thank you.
|
You are right, the probabilities are not the same, although I am a little puzzled by your case "$X=1$." The smallest possible value of the number of draws until we get our first match is $2$. But perhaps you are not counting the last draw.
So let $Y$ be the number of draws until the first match, including the draw that got us that match. It is easy to see that $P(Y=2)=\frac{1}{7}$. It doesn't matter what we first drew. The probability that the next draw matches it is $\frac{1}{7}$.
For $Y=3$, there are two possible patterns, which I will call $abb$ and $aba$ (where $a\ne b$). We don't care what the first draw is, call it $a$. The probability of getting something different ($b$) on the next draw is $\frac{6}{7}$. And given we got $a$ then $b$ on the first two draws, the probability of drawing $b$ on the third is $\frac{1}{6}$. So the probability of a pattern of type $abb$ is $\frac{6}{7}\cdot\frac{1}{6}$.
Similarly, given that we got the pattern $ab$ on the first two draws, the probability of getting $a$ on the third is $\frac{1}{6}$. Thus
$$P(Y=3)=\frac{6}{7}\cdot\frac{1}{7}+\frac{6}{7}\cdot\frac{1}{6}=\frac{2}{7}.$$
The calculation can be collapsed into one step. Given that we got something different from $a$ on the second draw, there are $2$ (out of $6$) tickets that will give us a match, so the probability is $\frac{6}{7}\cdot\frac{2}{6}$. The mistake in the calculation is in the last term of your product. You had $\frac{1}{6}$ there instead of the correct $\frac{2}{6}$. The $4\cdot\frac{2}{8}$ part was unnecessary but harmless.
|
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|
square squares with diagonals also squares The numbers reading across and down in these squares are square:
$\begin{array}{ccc}
1 & 4 & 4\\
4 &8&4\\
4&4&1
\end{array}$
$\begin{array}{ccc}
5&2&9\\
2&5&6\\
9&6&1
\end{array}$
$\begin{array}{cccc}
2&1&1&6\\
1&2&2&5\\
1&2&9&6\\
6&5&6&1
\end{array}$
Are there any such square squares where the diagonals are also squares? If not in base 10, is it possible in other bases?
|
Those matrices are certainly hard to find, I tried it a lot and found that there are no $4\times4$ or $3\times3$ matrices that do what you want in base $10$ or below. However watch this one in base $11$:
$$\begin{array}{ccc}
1 & 9 & 5\\
9 & 6 & 1\\
5 & 1 & 9
\end{array}$$
You have
$169_{11}=196=14^2$
$195_{11}=225=15^2$
$519_{11}=625=25^2$
$565_{11}=676=26^2$
$961_{11}=1156=34^2$
All colums, rows, diagonals when read in any way (colums and rows only down and right) are squares, woha ;-)
|
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|
How to get closed form from generating function? I have this generating function:
$$\frac{1}{2}\, \left( {\frac {1}{\sqrt {1-4\,z}}}-1 \right) \left( \,{
\frac {1-\sqrt {1-4\,z}}{2z}}-1 \right)$$
and I know that $\frac {1}{\sqrt {1-4\,z}}$ is the generating function for the sequence $\binom {2n} {n}$, and $\frac {1-\sqrt {1-4\,z}}{2z}$ is the generating function for the sequence $\frac {1}{n+1}\binom{2n} {n}$.
Now, I thought that I could substitute those in there, and where they multiply I'll use a summation like this:
$$\frac{1}{2}\left( 1-\frac{1}{n+1}\binom{2n} {n}-\binom{2n} {n} +
\sum_{k=0}^n \frac{1}{k+1} \binom{2k}{k}\binom{2(n-k)}{n-k}
\right)$$
Could this be right? It doesn't seem to work when I try in Maple. What else could I do?
I already know that the end sequence will be $\binom{2n-1}{n-2}$ if this can help...
|
Expanding the product, and after some algebra:
$$
g(z) = \frac{1}{2} \left(\frac{\sqrt{1-4 z}-1}{z}+\frac{1}{\sqrt{1-4 z}}+1\right)
$$
Using generalized binomial theorem:
$$
[z]^n g(z) = \frac{1}{2} \left( \delta_{n,0} + \binom{1/2}{n+1} (-4)^{n+1} + \binom{-1/2}{n} (-4)^n \right)
$$
Using
$$ \begin{eqnarray}
\binom{-1/2}{n} - 4 \binom{1/2}{n+1} &=& \frac{\Gamma(1/2)}{n! \Gamma(1/2-n)} - \frac{\Gamma(3/2)}{(n+1)! \Gamma(1/2-n)} \\
&=& (-1)^n (n-1) \frac{\Gamma(1/2+n) \Gamma(n+1)}{\Gamma(1/2) (n+1)!n!} \\
&=& (-1)^n (n-1) \frac{(2n)!}{4^n (n+1)!n!} \\
&=& (-1)^n \frac{(2n-1)!}{2 \cdot 4^{n-1} (n+1)!(n-2)!} = 2 \cdot \frac{(-1)^n}{4^n} \cdot \binom{2n-1}{n+1}
\end{eqnarray}
$$
Hence, using Iverson's bracket:
$$
[z]^n g(z) = \binom{2n-1}{n+1} \left[ n \ge 2 \right]
$$
|
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|
How to prove :If $p$ is prime greater than $3$ and $\gcd(a,24\cdot p)=1$ then $a^{p-1} \equiv 1 \pmod {24\cdot p}$? I want to prove following statement :
If $p$ is a prime number greater than $3$ and $\gcd(a,24\cdot p)=1$ then :
$a^{p-1} \equiv 1 \pmod {24\cdot p}$
Here is my attempt :
The Euler's totient function can be written in the form :
$n=p_1^{k_1}\cdot p_2^{k_2} \ldots \cdot p_r^{k_r} \Rightarrow \phi(n)=p_1^{k_1}\cdot\left(1-\frac{1}{p_1}\right)\cdot p_2^{k_2}\cdot\left(1-\frac{1}{p_2}\right)\ldots p_r^{k_r}\cdot \left(1-\frac{1}{p_r}\right)$
So,
$\phi(24 \cdot p)=2^3\cdot \left(1-\frac{1}{2}\right)\cdot3^1\cdot\left(1-\frac{1}{3}\right)\cdot p\cdot\left(1-\frac{1}{p}\right)=8\cdot(p-1)$
Euler's totient theorem states that :
if $\gcd(a,n)=1$ then $a^{\phi(n)} \equiv 1 \pmod n$
Therefore we may write :
$a^{\phi(n)}-1 \equiv 0 \pmod n \Rightarrow a^{\phi(24\cdot p)}-1=a^{8\cdot(p-1)}-1 \equiv 0 \pmod{24\cdot p} \Rightarrow$
$\Rightarrow \left(a^{p-1}\right)^8-1=(a^{p-1}-1)\cdot \displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \equiv 0\pmod{24\cdot p}$
So we may conclude :
$(a^{p-1}-1) \equiv 0 \pmod {24\cdot p}$ , or $\displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \equiv 0\pmod{24\cdot p}$
How can I prove that $\displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \not\equiv 0\pmod{24\cdot p}$ ?
|
Hint: The claim follows from proving the following facts separately: $a^{p-1}\equiv 1\pmod 8$, $a^{p-1}\equiv 1\pmod 3$, and $a^{p-1}\equiv 1\pmod p$.
|
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|
How to prove that $2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}$ I want to prove that$$2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}, x\geq 0$$
I have started from the arcsin part and I tried to end to the arctan one but I failed.
Can anyone help me solve it?
|
One way would be to notice that both functions have the same derivative and then find out what the "constant" is by plugging in $x=0$.
Here's another way. Look at
$$
\frac\pi2 - 2\arctan\sqrt{x} = 2\left( \frac\pi4 - \arctan\sqrt{x} \right) = 2\left( \arctan1-\arctan\sqrt{x} \right).
$$
Now remember the identity for the difference of two arctangents:
$$
\arctan u - \arctan v = \arctan\frac{u-v}{1+uv}.
$$
(This follows from the usual identity for the tangent of a sum.) The left side above becomes
$$
2\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}.
$$
The double-angle formula for the sine says $\sin(2u)=2\sin u\cos u$. Apply that:
$$
\sin\left(2\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}\right) = 2 \sin\left(\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\cos\left(\arctan\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)
$$
Now remember that $\sin(\arctan u) = \dfrac{u}{\sqrt{1+u^2}}$ and $\cos(\arctan u) = \dfrac{1}{\sqrt{1+u^2}}$
Then use algebra:
$$
2\cdot\frac{\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)}{\sqrt{1+\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^2}}\cdot \frac{1}{\sqrt{1+\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)^2}} = \frac{1-x}{1+x}.
$$
|
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|
Find a closed term for $f(n) = n + 2 f(n-1)$, $f(1)=1$ I cannot help myself, but I don't get the closed term for: $f(n) = n + 2 f(n-1)$, where f(1) = 1. I tried to find the pattern when looking at some iterations, and I think I see the pattern very clearly:
$...(6 + 2 ( 5 + 2 ( 4 + 2 ( 3 +2 ( 2 \cdot 1 ) ) ) ) ... )$
It's always n, reduced by the iteration plus two times the next iteration.
Any hints or what construct I should use to find the term?
|
Just for fun, here is a solution using generating functions.
Let $F(x)=\sum_{n=1}^\infty f(n)x^n$. Then,
$$\begin{align*}
F(x)&=\left(\sum_{n=2}^\infty (n+2f(n-1))x^n\right)+x\\
&=\left(\frac{x}{(1-x)^2}-x+2x\sum_{n=1}^\infty f(n)x^n\right)+x\\
&=\frac{x}{(1-x)^2}+2xF(x)
\end{align*}$$
So,
$$F(x)=\frac{x}{(1-x)^2(1-2x)}=\frac{1}{x-1}-\frac{1}{(1-x)^2}+2\left(\frac{1}{1-2x}\right)$$
where the far right hand side is the partial fraction decomposition.
Now,
$$\frac{1}{x-1}=-\sum_{n=0}^\infty x^n$$
$$\frac{1}{(1-x)^2}=\sum_{n=0}^\infty (n+1)x^n$$
$$2\left(\frac{1}{1-2x}\right)=2\sum_{n=0}^\infty (2x)^n$$
Hence,
$$F(x)=\sum_{n=0}^\infty (-1-(n+1)+2^{n+1})x^n$$
So, since the coefficient of $x^n$ is $f(n)$, we get $f(n)=2^{n+1}-n-2$.
|
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|
How to remember the trigonometric identities I have a test tomorrow and I am having trouble remembering those pesky trigonometrical identities (such as $1-\cos x=2\sin^2(\frac{x}{2})$ )
Do you guys have any tips on how I can remember these?
Thanks :)
|
My favourite trick: I don't remember any of them. :-) The only thing I have in mind is that this matrix
$$
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
$$
rotates vectors in the plane by an angle $\theta$ and matrix multiplication is the same as composition. Hence, you have identities like
$$
\begin{pmatrix}
\cos(2\theta) & -\sin(2\theta) \\
\sin(2\theta) & \cos(2\theta)
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
$$
from which it follows
$$
\cos(2\theta) = \cos^2\theta - \sin^2\theta
$$
and
$$
\sin(2\theta) = 2\sin\theta\cos\theta \ .
$$
Alternatively, as yoyo says, you could use Euler's identity,
$$
e^{i\theta} = \cos\theta + i \sin\theta
$$
to find, for instance, that
$$
\cos(\theta + \phi) + i\sin(\theta + \phi) = e^{i(\theta + \phi)} = e^{i\theta}e^{i\phi} = (\cos\theta + i\sin\theta) (\cos\phi + i\sin\phi) \ .
$$
Hence,
$$
\cos(\theta + \phi) = \cos\theta\cos\phi - \sin\theta\sin\phi
$$
and
$$
\sin(\theta + \phi) = \sin\theta\cos\phi + \cos\theta\sin\phi \ .
$$
|
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|
On computing: $ \gcd \left({2n \choose 1}, {2n \choose 3},\cdots, {2n \choose 2n-1}\right)$ I would like to calculate
$$ d=\gcd \left({2n \choose 1}, {2n \choose 3},\cdots, {2n \choose 2n-1}\right) $$
We have:
$$ \sum_{k=0}^{n-1}{2n \choose 2k+1}=2^{2n-1} $$
$$ d=2^k, 0\leq k\leq2n-1 $$
...
Any idea?
|
Let $q = 2^i$ where $2^i | 2n$ and $2^{i+1} \not|2n$. Claim: $d=q$.
First we'll show that each term ${2n \choose 2k+1}$, for $0 \leq k \leq n-1$ has $q$ as a factor. Consider,
$$\begin{align*}
{2n \choose 2k+1} &= \frac{(2n)(2n-1)(2n-2) \cdots (2n- [2k+1] + 1)}{(2k+1)(2k)(2k-1)\cdots(1)}\\
&= \frac{2n}{2k+1} \cdot {2n-1 \choose 2k}\\
&= \frac{2n{2n-1 \choose 2k}}{2k+1}
\end{align*}$$
Since $k \leq n-1$, $2k < 2n-1$, and the number ${2n-1 \choose 2k}$ is a nonzero integer. Now since ${2n \choose 2k+1}$ is an integer, we know that $2k+1$ divides the numerator. There are no factors of $2$ in $2k+1$, therefore,
$q$ must survive to be a factor of ${2n \choose 2k+1}$.
Second we show that $d \leq q$. We know at least that $q$ and $d$ share the same highest power of 2, since one of the terms is ${2n \choose 1} = 2n$.
Now as Andre points out in the comments, this means we're done at this point. Since OP proved that $d$ is a power of 2, we have $d=q$.
|
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|
How to find the derivative of $f(x)$? This is not a homework problem.
How to find $f'(x)$ if $f(x)=2x^{1/2}\log(2)$?
Thanks in advance for your help! I just can't figure it out...
What rule should I use?
|
Note that $\frac{d}{dx} (c \cdot f(x)) = c \cdot f'(x)$. Hence you just want to find the derivative of $f(x) = \sqrt{x}$. Using the definition of the derivative we have
\begin{align*}
f'(x) & = \lim_{h \to 0} \: \frac{f(x+h)- f(x)}{h} \\ &= \lim_{h \to 0} \: \frac{\sqrt{x+h} - \sqrt{x}}{h} \\ &= \lim_{h \to 0} \: \frac{\sqrt{x+h}-\sqrt{x}}{h} \times \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ &=\lim_{h \to 0} \: \frac{x+h - x}{h \cdot \bigl(\sqrt{x+h} + \sqrt{x}\:\bigr)} \\ &= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2 \cdot \sqrt{x}}
\end{align*}
Use the above rule and see that $ \displaystyle f'(x) = 2 \cdot \log(2) \cdot \frac{1}{2 \cdot \sqrt{x}}$
|
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|
Combinatorial interpretation of sum of squares, cubes Consider the sum of the first $n$ integers:
$$\sum_{i=1}^n\,i=\frac{n(n+1)}{2}=\binom{n+1}{2}$$
This makes the following bit of combinatorial sense. Imagine the set $\{*,1,2,\ldots,n\}$. We can choose two from this set, order them in decreasing order and thereby obtain a point in $\mathbb{N}^2$. We interpret $(i,*)$ as $(i,i)$. These points give a clear graphical representation of $1+2+\cdots+n$:
$$
\begin{matrix}
&&&\circ\\
&&\circ&\circ\\
&\circ&\circ&\circ\\
\circ&\circ&\circ&\circ\\
\end{matrix}
$$
Similar identities are:
$$\sum_{i=1}^n\,i^2=\frac{1}{4}\binom{2n+2}{3}=\frac{1}{2n-1}\binom{2n+2}{4}=\frac{1}{2n+3}\binom{2n+3}{4}$$
$$\sum_{i=1}^n\,i^3=\binom{n+1}{2}^2$$
I am aware of geometric explanations of these identities, but not combinatorial ones similar to the above explanation for summing first powers that make direct use of the "choosing" interpretation of the binomial coefficient. Can anyone offer combinatorial proofs of these?
|
We give a combinatorial interpretation of the formula
$$
2^2+4^2+6^2+\cdots +(2n)^2=\binom{2n+2}{3} \qquad\qquad (1)
$$
for the sum of the first $n$ even squares.
There are $\binom{2n+2}{3}$ ways to choose $3$ numbers from the numbers $1, 2, \dots, 2n+2$. We organize and count the choices in another way.
Maybe the largest number chosen is $2k+1$ or $2k+2$. If it is $2k+1$, the other two can be chosen in $\binom{2k}{2}$ ways, and if it is $2k+2$, then the other two can be chosen in $\binom{2k+1}{2}$ ways. The total is
$$\frac{(2k)(2k-1)}{2} +\frac{(2k+1)(2k)}{2}\quad\text{that is,}\quad (2k)^2.\qquad\qquad(2)$$
Take $k=1, 2, 3, \dots, n$. By Formula $(2)$, the number of ways of choosing $3$ numbers from $1, 2, \dots, 2n+2$ is
$$
2^2+4^2+6^2+\cdots +(2n)^2.
$$
The above argument was not purely bijective, because of the ``calculation'' in Formula $(2)$. But there is a standard workaround that produces a purely bijective proof of the fact that
$$
\binom{m}{2} +\binom{m+1}{2}=m^2.
$$
The geometric idea is that the sum of two consecutive triangular numbers is a square. More formally, let $M$ be the collection $\{1,2,3,\dots,m\}$. Then the number of ordered pairs $(a,b)$ with $a$ and $b$ in $M$ is $m^2$. These ordered pairs are of two types: (i) the ones with $a<b$ and (ii) the ones with $a \ge b$. The number of ordered pairs $(a,b)$ with $a<b$ is just $\binom{m}{2}$. The number of ordered pairs $(a,b)$ with $a \ge b$ is the same as the number of ordered pairs $(b,a+1)$ with $b<a+1$, and this is $\binom{m+1}{2}$.
Comment:
It follows by "algebra" from Formula $(1)$ that
$$
1^2+2^2+\cdots +n^2=\frac{1}{4}\binom{2n+2}{3}. \qquad\qquad(3)
$$
(The algebraic division by $4$ can be bypassed by counting suitable equivalence classes, giving a pristinely bijective proof of $(3)$.)
When I first saw the usual expression $\dfrac{(n)(n+1)(2n+1)}{6}$ for the sum of the first $n$ squares, I remember thinking that the $2n+1$ somehow looks as if it does not fit in. But it does, it is actually $n$ and $n+1$ that are strange! We can think of $(n)(n+1)(2n+1)$ as ``really'' being $(1/4)[(2n)(2n+1)(2n+2)]$.
|
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|
Trigonometric expressions possible pattern While solving trigonometric problems I've noticed possible pattern that reminded me of Fibonacci numbers and Pascal triangle. So I tried to find next "element" of this pattern (8 degree exponent ) and failed to do so. So can someone please tell me is there a pattern or just "coincidence".
Possible pattern :
$\sin^2x + \cos^2x = 1 $
$\sin^4x + \cos^4x = 1 - 2\sin^2x\cos^2x$
$\sin^6x + \cos^6x = 1 - 3\sin^2x\cos^2x$
|
There is a pattern but I do not think it is as simple as what the first three terms might suggest.
Let me use the abbreviation $s := \sin x$ and $c := \cos x$. In this notation, we have
$$
(s^6 + c^6) = (s^6 + c^6)(s^2 + c^2) = (s^8 + c^8) + s^2c^2(s^4+c^4).
$$
Now plugging in the expressions you obtained for $s^6+c^6$ and $s^4+c^4$, we get
$$
1 - 3s^2c^2 = (s^8+c^8)+s^2c^2(1-2s^2c^2),
$$
or:
$$
s^8+ c^8 = 1 - 4s^2c^2 + 2s^4c^4.
$$
Alternate derivation: Since $s^2 + c^2 = 1$, let $s^2 = \frac{1}{2} + u$ and $c^2 = \frac12 - u$ for some $u$. Then
$$
s^2 c^2 = \frac14 - u^2. \tag{$\dagger$}
$$
Now calculating $s^8+c^8$ is straightforward, if a bit involved:
$$
\begin{align*}
s^8 + c^8
&= \left(\frac12 + u\right)^4 + \left(\frac12 - u\right)^4
\\\\ &= 2 \left( \frac{1}{2^4} + u^4 + 6 \cdot \frac{1}{2^2} \cdot u^2 \right)
\\\\ &= \frac{1}{8} +2 u^4 + 3 u^2
\\\\ &\stackrel{(\dagger)}{=} \frac{1}{8} +2 \left( \frac14 - s^2 c^2 \right)^2 + 3 \left( \frac14 - s^2 c^2 \right)
\\\\ &= \frac18 + 2\cdot \frac{1}{4^2} + \frac34 +2 s^4 c^4 - 2 \cdot 2 \cdot \frac14 \cdot s^2 c^2 - 3 s^2 c^2
\\\\ &= 1 + 2 s^4 c^4 - 4s^2 c^2 .
\end{align*}
$$
Although this seems longer, the second method generalises easily for $s^{2n}+c^{2n}$ for all $n$.
What does $\sin^{2n} x + \cos^{2n} x$ look like?
It is a standard fact that any symmetric polynomial in $\lambda_1$ and $\lambda_2$ can be expressed in terms of the elementary symmetric polynomials, which, in this case, is just $\lambda_1 + \lambda_2$ and $\lambda_1 \lambda_2$. Now for $\lambda_1 = s^2 = \sin^2 x$ and $\lambda_2 = c^2 = \cos^2 x$, we have $\lambda_1 + \lambda_2 = 1$. That is, every symmetric polynomial in $s = \sin x$ and $c = \cos x$ with the restriction that every term has an even degree in both variables can be expressed as a polynomial in $\sin x\cos x$.
For the special case of $s^{2n} + c^{2n}$, I already explained how to get such a representation:
$$
s^{2n} + c^{2n}
= \left( \frac12 + u \right)^n + \left( \frac12 - u \right)^n.
$$
Expanding using the binomial theorem, we find that all the odd powers of $x$ cancel and the even powers add up, so that we have an even polynomial in $x$. The degree of this polynomial is exactly $2 \lfloor n/2 \rfloor$.
At the next step, we replace $u^2$ by $\frac14 - s^2c^2$, so that we again get an even polynomial -- this time in $s \cdot c$ -- of the same degree, namely $2 \lfloor n/2 \rfloor$. In other words, we get a polynomial in $s^2 c^2$ of degree $\lfloor n / 2 \rfloor$. (However, the actual polynomials themselves do not seem very interesting.)
In general, one should not expect much more simplification. As Henning notes below, for $n =2$ and $n=3$, the expression $\lfloor n / 2 \rfloor$ is just $1$, which is why we got a polynomial of containing only a constant term and a $\sin^2 x \cos^2 x$ term. This pretty pattern disappears even for $n=4$.
|
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|
Converting $\tanh^{-1}{x}$ to an expression involving the natural logarithm I know how to convert $\sinh^{-1}{x}$ and $\cosh^{-1}{x}$ to $\ln{|x+\sqrt{x^2 \pm 1}|}$, but for some reason I am struggling to do the same for the following statement:
$$\tanh^{-1}{\frac{x}{2}}$$
Can someone please show me how to convert it to a $\ln$ form? thanks!
|
$$\tanh x=\frac{e^{2x}-1}{e^{2x}+1} \Rightarrow \tanh \frac{x}{2}=\frac{e^{x}-1}{e^{x}+1} $$ In order to find $artanh\frac{x}{2}$ we have to solve following equation :
$$\frac{x}{2}= \frac {e^{2artanh \frac{x}{2}}-1}{e^{2artanh \frac{x}{2}}+1} \Rightarrow x \cdot e^{2artanh \frac{x}{2}}+x=2e^{2artanh \frac{x}{2}}-2 \Rightarrow$$
$$x+2=(2-x) \cdot e^{2artanh \frac{x}{2}} \Rightarrow e^{2artanh \frac{x}{2}}=\frac{2+x}{2-x} \Rightarrow artanh \frac{x}{2} =\frac{1}{2} \ln \left(\frac{2+x}{2-x}\right) ; |x| < 2$$
|
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|
Limit $\lim_{n\to+\infty} (1-\frac1{2^2})(1-\frac1{3^2})\cdot \cdots \cdot(1-\frac{1}{n^2})$ and series $\sum_{n=2}^{\infty} \ln(1-\frac1{n^2})$
Possible Duplicate:
Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$
Compute:
\begin{align*}
\lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})
\end{align*}
Well, I do so:
$$\lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})=\lim_{n\to+\infty}\prod_{j=2}^n (1-\frac{1}{j^2})$$
let $$a_n = \prod_{j=2}^n (1-\frac{1}{j^2})\quad\Rightarrow\quad \ln a_n = \ln\left(\prod_{j=2}^n (1-\frac{1}{j^2})\right)$$
so:
$$\ln\left(\prod_{j=2}^n (1-\frac{1}{j^2})\right)=\sum_{j=2}^{\infty} \ln (1-\frac{1}{j^2})$$
consider
$$\sum_{n=2}^{\infty} \ln(1-\frac{1}{n^2})$$
but as you study this series??
|
Actually there is a neat form for the partial products.
Let
$$f(n)=(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})=\prod _{k=2}^n \left(1-\frac{1}{k^2}\right)$$
We show by induction that $f(n)=\frac{n+1}{2 n}$. $$f(n)=\left(1-\frac{1}{n^2}\right)f(n-1)=\frac{n^2-1}{n^2}\cdot\frac{n}{2(n-1)}=\frac{n+1}{2n}$$
Therefore $\lim_{n\to+\infty}f(n)=\lim_{n\to+\infty}\frac{n+1}{2 n}=\lim_{n\to+\infty}(\frac{1}{2}+\frac{1}{2n})=\frac{1}{2}$
|
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|
Integration - Partial Fraction Decomposition I was wondering if someone would be kind enough to walk me through the logic used in solving the following integral. I have been to class, and have read the section (9.4 of Swokowski's Classic), and have studied the answer in the solution manual, but I can't quite seem to make sense of the rules posed (p.474, Swokowski's Classic) for the decomposition.
$$
\int\frac{x^2+3x+1}{x^4+5x^2+4}dx
$$
Factors to:
$$
\int\frac{x^2+3x+1}{(x^2+4)(x^2+1)}dx
$$
And this is where I get completely lost. I can do simple ones, such as
$$
\int\frac{x+16}{x^2+2x-8}
$$
where they reduce to
$$
\frac{A}{x+4} + \frac{B}{x-2}
$$
But the solution manual suggests that A and B should be Ax+B and Cx+D, referring to the aforementioned rule, and I'm quite confused.
Thank you very much!
|
Procedure of decomposition
So:
$$\frac{x^2+3x+1}{(x^2+4)(x^2+1)} = \frac{ax+b}{x^2+4}+\frac{cx+d}{x^2+1} \Rightarrow$$
$$ \Rightarrow x^2+3x+1 =(x^2+1)(ax+b)+(x^2+4)(cx+d) \Rightarrow$$
$$\Rightarrow x^2+3x+1 =(a+c)x^3+(b+d)x^2+(a+4c)x+(b+4d)$$
Therefore , you have to solve following system of equations :
$\begin{cases}
a+c=0 \\
b+d=1 \\
a+4c=3 \\
b+4d=1
\end{cases}$
After you find coefficients : $a,b,c,d$ use sum of integrals :
$$\int\frac{ax+b}{x^2+4} \,dx =\int\frac{ax}{x^2+4} \,dx+\int\frac{b}{x^2+4} \,dx$$
$$\int\frac{cx+d}{x^2+1} \,dx =\int\frac{cx}{x^2+1} \,dx+\int\frac{d}{x^2+1} \,dx$$
|
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"url": "https://math.stackexchange.com/questions/103625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
how to work out a closed form of a sequence
Consider the following linear recurrence sequence.
$x_1 = 11$, $x_{n+1} = -0.8x_n + 9,\quad n = 1,2,3, \ldots.$
Find a closed form for this sequence.
|
To get an idea of what the closed form might look like, let's iterate the relation a few times:
$$
\begin{align}
x_n
&=9-.8x_{n-1}\\
&=9-.8(9-.8x_{n-2})\\
&=9-.8\cdot9+.8^2x_{n-2}\\
&=9-.8\cdot9+.8^2(9-.8x_{n-3})\\
&=9-.8\cdot9+.8^2\cdot9-.8^3x_{n-3}\tag{1}
\end{align}
$$
Looking at $(1)$, it appears that
$$
x_n=9\frac{1-(-.8)^k}{1+.8}+(-.8)^kx_{n-k}\tag{2}
$$
Equation $(2)$ can be verified using induction.
Verification: The case $k=1$ is just the given recursion: $x_n=9-.8x_{n-1}$.
Suppose $(2)$ is true for some $k$, then
$$
\begin{align}
x_n
&=9\frac{1-(-.8)^k}{1+.8}+(-.8)^kx_{n-k}\\
&=9\frac{1-(-.8)^k}{1+.8}+(-.8)^k(9-.8x_{n-k-1})\\
&=9\frac{1-(-.8)^k}{1+.8}+9(-.8)^k+(-.8)^{k+1}x_{n-k-1}\\
&=9\frac{1-(-.8)^{k+1}}{1+.8}+(-.8)^{k+1}x_{n-k-1}\\
\end{align}
$$
So $(2)$ is true for $k+1$.
Using $x_1=11$ and $k=n-1$, $(2)$ becomes
$$
\begin{align}
x_n
&=5(1-(-.8)^{n-1})+(-.8)^{n-1}11\\
&=5+6(-.8)^{n-1}\tag{3}
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/104691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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|
Probability of All Distinct Faces When Six Dice Are Rolled If six fair dice are rolled what is probability that each of the six numbers will appear exactly once?
|
Imagine you throw one after the other. You consider a throw as a success if the number is different from all previous numbers. You start with one. This is always a succes so $P(\text{first}) = 1 = \frac{6}{6}$. Your second throw is a success if one of the remaining $5$ numbers shows, so $P(\text{second}) = \frac{5}{6}$. And so on. Since all the throws are independent, the total probability is the product of all separate probabilities:
$P(\text{all numbers are different}) = \frac{6}{6} \cdot \frac{5}{6} \cdot \frac{4}{6} \cdot \frac{3}{6} \cdot \frac{2}{6} \cdot \frac{1}{6}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/105300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
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|
Does $x^2\equiv x\pmod p$ imply $x^2\equiv x\pmod {p^n}$ for all $n$? Suppose $x^2\equiv x\pmod p$ where $p$ is a prime, then is it generally true that $x^2\equiv x\pmod {p^n}$ for any natural number $n$? And are they the only solutions?
|
Way 1: Let us compute a bit. Let $p=2$. Note that $x^2\equiv x \pmod 2$ for any $x$. Is it true that always $x^2\equiv x\pmod 4$? No, let $x=2$.
Way 2: We do more work, but will get a lot more information. Rewrite the congruence $x^2\equiv x \pmod p$ as $x^2-x\equiv 0 \pmod p$, and then as $x(x-1)\equiv 0\pmod p$.
This says that the product of $x$ and $x-1$ is divisible by $p$. A product $ab$ is divisible by the prime $p$ if and only if $p$ divides $a$ or $p$ divides $b$, or both.
So $x(x-1)$ is divisible by $p$ if and only if $x$ is divisible by $p$ or $x-1$ is divisible by $p$, that is, if and only if $x\equiv 0 \pmod p$ or $x\equiv 1 \pmod p$.
Now does this force $x(x-1) \equiv 0 \pmod {p^2}$? No, for example we could take $x=p$. Certainly it is not true that $p(p-1)$ is divisible by $p^2$. We could also take $x=2p$, or $x=3p$, and so on up to $x=(p-1)p$. Or we could put the badness in the $x-1$ part, by putting $x=1+p$, or $x=1+2p$, and so on up to $x=1+(p-1)p$. We get a total of $2(p-1)$ numbers $x$ between $0$ and $p^2-1$ such that $x^2\equiv x \pmod p$ but $x^2\not\equiv x \pmod{p^2}$.
We can similarly identify the numbers $x$ between $0$ and $p^n-1$ such that $x(x-1)\equiv 0 \pmod p$ but $x(x-1)\not \equiv 0 \pmod{p^n}$. Again, they are of two types: (i) the numbers $p$, $2p$, and so on up to $p(p^{n-1}-1)$ and (ii) the numbers obtained by adding $1$ to numbers in list (i).
|
{
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"url": "https://math.stackexchange.com/questions/109961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
What is the probability that 5 digit number divisible by 6? The main constraint is that each digit can only take digits from $\{1, 2, 3, 4, 5\}$. So the sample space will be $5^{5}$.
What is the probability that a random number taken from this sample space will be divisible by $6$?
Thanks.
|
$$\color{red}{416/3125}=0.13312.
$$
The last digit must be $2$ or $4$, this happens with probability $2/5$.
The sum of the four other digits must be $\pm1\pmod{3}$, according to the last digit being $2$ or $4$. Since both events have the same probability, the answer is $2/5$ times the probability that the sum $s$ of four digits is $1\pmod{3}$, that is, $s=-2$ or $s=+1$ or $s=+4$.
$s=+4$ corresponds to $+1,+1,+1,+1$, with probability $2^4/5^4$.
$s=+1$ corresponds to $0,0,0,+1$, or $0,+1,+1,-1$ in whatever order. In the first case, one must place the $+1$, thus $4$ cases, with probability $2/5^4$ each. In the second case, one must place the $0$ and the $-1$, thus $12$ cases, with probability $2^3/5^4$ each.
$s=-2$ corresponds to $+1,-1,-1,-1$, thus $4$ cases, with probability $2^4/5^4$ each, or to $0,0,-1,-1$, thus $6$ cases, with probability $2^2/5^4$ each.
Summing up, the answer is $(2/5)\cdot(2^4+4\cdot2+12\cdot2^3+4\cdot2^4+6\cdot2^2)/5^4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/110324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Integral $\int\csc^3{x} \ dx$ I found these step which explain how to integrate $\csc^3{x} \ dx$. I understand everything, except the step I highlighted below.
How did we go from:
$$\int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx%$$
to
$$\int \frac{d(-\cot x + \csc x)}{-\cot x + \csc x} \quad?$$
Thank you for your time!
$$
\int \csc^3 x\,dx = \int\csc^2x \csc x\,dx$$
To integrate by parts, let $dv = \csc^2x$ and $u=\csc x$. Then $v=-\cot x$ and $du = -\cot x \csc x \,dx$.
Integrating by parts, we have:
$$\begin{align*}
\int\csc^2 x \csc x \,dx &= -\cot x \csc x - \int(-\cot x)(-\cot x\csc x\,dx)\\
&= -\cot x \csc x - \int \cot^2 x \csc x\,dx\\
&= -\cot x\csc x - \int(\csc^2x - 1)\csc x\,dx &\text{(since }\cot^2 x = \csc^2-1\text{)}\\
&= -\cot x \csc x - \int(\csc^3 x - \csc x)\,dx\\
&= -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx
\end{align*}$$
From
$$\int \csc^3 x\,dx = -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx$$
we obtain
$$\begin{align*}
\int\csc^3x\,dx + \int\csc^3 x\,dx &= -\cot x \csc x + \int\csc x\,dx\\
2\int\csc^3 x\,dx &= -\cot x\csc x + \int\csc x\,dx\\
\int\csc^3x\,dx &= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\csc x\,dx\\
&=-\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\frac{\csc x(\csc x - \cot x)}{\csc x - \cot x}\,dx\\
&= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{\csc^2 x - \csc x\cot x}{\csc x - \cot x}\,dx\\
&= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{d(-\cot x+\csc x)}{-\cot x +\csc x}\\
&= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\ln|\csc x - \cot x|+ C
\end{align*}$$
|
The question seems to be why the following are equal:
$$(\csc^2 x - \csc x \cot x)\,dx = d(-\cot x + \csc x)$$
The answer is that
$$
\frac{d}{dx} \cot x = -\csc^2 x\quad \text{ and }\quad\frac{d}{dx} \csc x = -\csc x \cot x.
$$
The quotient rule for derivatives can establish both of these identities if you know how to differentiate the sine and cosine and some simple trigonometric identities.
$$
\begin{align}
\frac{d}{dx} \cot x & = \frac{d}{dx} \frac{\cos x}{\sin x} = \frac{(\sin x)\frac{d}{dx}\cos x - (\cos x) \frac{d}{dx} \sin x}{\sin^2 x} \\ \\ \\
& = \frac{(\sin x)(-\sin x) - (\cos x)(\cos x)}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x.
\end{align}
$$
And similarly,
$$
\frac{d}{dx} \csc x = \frac{d}{dx} \frac{1}{\sin x} = \frac{- \frac{d}{dx} \sin x}{\sin^2 x} = \frac{-\cos x}{\sin^2 x} = -\;\frac{1}{\sin x}\frac{\cos x}{\sin x} = -\csc x \cot x.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/112918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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|
Integrate using Partial Fraction decomposition, completing the square The given problem is $\int{x\over x^3-1}dx$.
I know this equals
$${1\over3}\int {1\over x-1}-{x-1\over x^2+x+1}dx,$$
which can be separated into
$${1\over3}\int {1\over x-1}dx - {1\over3}\int{x+(1/2)-(3/2)\over x^2+x+1}dx.$$ This can further be separated into $${1\over3}\int{1\over x-1}dx - {1\over3}\int{x+(1/2)\over x^2+x+1}dx + {1\over2}\int{1\over(x+(1/2))^2+(3/4)}dx.$$
I know the integral of ${1\over3}\int {1\over x-1}dx$ is $(1/3)\ln(x+1)+C$ where $C$ is an arbitrary constant. Using u-subsitution, where $u=x^2+x+1$ and $du=(2x+1)dx$ and $(1/2)du=(x+(1/2))dx$, I know the integral of $-{1\over3}\int{ x+(1/2)\over x^2+x+1 }dx$ is $(1/6)\ln(x^2+x+1)+C$. I need to get the last part, $${1\over2}\int{1\over(x+(1/2))^2+(3/4)}dx$$ to be some form of arctan. I can use u-substitution where $u=x+(1/2)$ and $du=dx$, but I don't know where to go from there.
|
Use the substitution
$$\frac{\sqrt{3}}{2}u=x+\frac{1}{2}.$$
Or else, if you want to do it in two steps, make the substitution $u=x+\frac{1}{2}$.
You will end up with an expression that includes $u^2+\frac{3}{4}$. Then let $u=\frac{\sqrt{3}}{2}v$. The substitution that I proposed is a little faster.
Remark: Suppose that we want to find the integral
$$\int\frac{dx}{x^2+k},$$
where $k$ is a positive constant. Write $k=a^2$, where $a$ is positive. Then $a=\sqrt{k}$. We are interested in
$$\int\frac{dx}{x^2+a^2}.$$
Let $x=aw$ (or equivalently, $w=\frac{x}{a}$). Then $dx=a\,dw$. Substituting, we get
$$\int \frac{a\,dw}{a^2w^2+a^2}, \quad\text{which is}\quad \int\frac{1}{a}\frac{dw}{w^2+1}.$$
More or less the same idea works, for example, when we want to integrate something that involves $\sqrt{k-x^2}$, where $k$ is positive. Write $k=a^2$, where we choose $a$ positive. So we are interested in $\sqrt{a^2-x^2}$. Make the substitution $u=a\sin\theta$. Or else, make the preliminary substitution $x=aw$. Then our expression becomes $\sqrt{a^2-a^2w^2}$, which is equal to $a\sqrt{1-w^2}$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$ The following expression is given:
$$\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$$
Simplify it, knowing that $x+y+z=0$.
|
Exploit the innate symmetry! Using Newton's identities to rewrite the power sums as elementary symmetric functions is very simple because $\rm\:e_1 = x+y+z = 0\:$ kills many terms.
Write $\rm\ \ c = e_2 = xy + yz + zx,\ \ \ d = e_3 = xyz,\ \ \ p_k =\: x^k + y^k + z^k.$
$\rm\qquad\qquad p_1\ =\ e_1 = 0$
$\rm\qquad\qquad p_2\ = {-}2\: c$
$\rm\qquad\qquad p_3\ =\ \ \ 3\: d$
$\rm\qquad\qquad p_4\ = - c\: p_2\ =\ 2\: c^2$
$\rm\qquad\qquad p_5\ = -c\: p_3 +\: d\: p_2\ = {-}5\: c\: d$
$\rm\qquad\qquad p_7\ = -c\: p_5 +\: d\: p_4\ =\ 7\: c^2 d$
Hence $\rm\displaystyle\ \frac{p_7}{p_4 d}\: =\: \frac{7\: c^2\: d}{2\: c^2\: d}\: =\: \frac{7}{2}$
|
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"url": "https://math.stackexchange.com/questions/115520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Find the minimum value.
Find the minimum value of $4^x + 4^{1-x}$ , $x\in\mathbb{R}$.
In this I used the property that $a + \frac{1}{a}\geq 2$.
So I begin with
$$ 4^x + \left(\frac{1}{4}\right)^x + 3\left(\frac{1}{4}\right)^x \geq 2 + 3\left(\frac{1}{4}\right)^x$$
So I think the minimum value be between 2 to 3.
But the answer is 4.
Thanks in advance.
|
$$4^{x} + 4^{1-x} = (\sqrt{4^x} - \frac{2}{\sqrt{4^x}})^2 + 4 \ge 4$$
with equality occuring when $\sqrt{4^x} = \frac{2}{\sqrt{4^x}}$ and so $x = \frac{1}{2}$.
|
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"url": "https://math.stackexchange.com/questions/115579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If we know the GCD and LCM of two integers, can we determine the possible values of these two integers? I know that $\gcd(a,b) \cdot \operatorname{lcm}(a,b) = ab$, so if we know $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$ and we want to find out $a$ and $b$, besides factoring $ab$ and find possible values, can we find these two integers faster?
|
Given gcd(a,b) and lcm(a,b), the possible values for a (which are the same as the possible values for b, by symmetry) are the unitary divisors of lcm(a,b)/gcd(a,b) multiplied by gcd(a,b). So suppose gcd(a,b) = 3 and lcm(a,b) = 6300. Factoring 6300/3 gives $2^2\cdot3\cdot5^2\cdot7$ and so the possible values are the members of
$\{3,2^2\cdot3,3^2,2^2\cdot3^2,3\cdot5^2,2^2\cdot3\cdot5^2,3^2\cdot5^2,2^2\cdot3^2\cdot5^2,3\cdot7,2^2\cdot3\cdot7,3^2\cdot7,2^2\cdot3^2\cdot7,3\cdot5^2\cdot7,2^2\cdot3\cdot5^2\cdot7,3^2\cdot5^2\cdot7,2^2\cdot3^2\cdot5^2\cdot7\}$
(taking, in each case, all factors of a prime together, then multiplying by 3).
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/116014",
"timestamp": "2023-03-29T00:00:00",
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|
Help with the line integral $\int x^2dS$ where C is a upper-half of a circle $x^2+y^2=r^2$ $$\int_C x^2dS$$ where C: $$x^2+y^2=r^2$$
So, $$ x=r\cos(\theta) $$ $$ y=r\sin(\theta) $$ $$ 0 \leq \theta \leq \pi $$
How would arc length of this curve go? $$ dS = \sqrt{\left(\frac{dr}{d\theta}\right)^2+r^2}d\theta $$ What should I put in $\left(\frac{dr}{d\theta}\right)^2$ and $r^2$?
|
$r$ is constant and $> 0$ in a circle. Therefore:
$
\frac{dr}{d\theta}=0
$
And
$
ds = \sqrt{(0)^2+(r)^2} d\theta = r d\theta
$
From there, your integral should be straightforward.
Another way to look at it:
\begin{align}
ds &= \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \ d\theta \\
\frac{dx}{d\theta} &= -r\sin{\theta} \\
\frac{dy}{d\theta} &= r\cos{\theta} \\
\Rightarrow ds &= \sqrt{(-r\sin{\theta})^2 + (r\cos{\theta})^2}d\theta = r d\theta \\
\end{align}
|
{
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"url": "https://math.stackexchange.com/questions/116337",
"timestamp": "2023-03-29T00:00:00",
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|
interesting square of log sin integral I ran across this challenging log sin integral and am wondering what may be a good approach.
$$
\int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}(2\cos(x))dx=\frac{11{{\pi}^{5}}}{1440}
$$
This looks like it may be able to be connected to the digamma or incomplete beta function somehow.
I tried using the identity
$$
\cos(x)=\frac{e^{ix}+e^{-ix}}{2},
$$
but that square threw a wrench in my plans.
Maybe write it as
$$
(x\ln(2\cos(x)))^{2}=\ln^{2}(2)x^{2}+2\ln(2)x^{2}\ln(\cos(x))+x^{2}\ln^{2}(\cos(x))
$$ and expand. I doubt if that does any good though.
Does anyone know of a clever way to approach this? It looks like a fun one if I knew a good starting point.
Is there an identity that goes with
$$
(2\cos(t))^{a}.
$$
If
$$
t^{2}(2\cos(t))^{a}
$$
were differentiated twice w.r.t a, then we would have
$$
(2\cos(t))^{a}t^{2}\ln^{2}(2\cos(t))
$$
Maybe the trig form of Beta which would give
$$
2^{a}\cdot B(1/2, \frac{a+1}{2}).
$$
Then, differentiate this twice and it would somehow relate to the digamma?
$$
\frac{1}{2}B(1/2,\frac{a+1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{a+1}{2})}{2\Gamma(\frac{a+2}{2})}.
$$
Differentiate twice and obtain some sort of Psi relation.
Cheers Everyone.
|
As I remarked in a comment, I have spent a while trying to derive this identity by means of contour integration. Though I have not succeeded at that, (I take great pleasure in striking that out; see my other answer below!) I have found some related identities in my search which surprised me quite a bit, and I think they're worth sharing. For example, using the integral you asked about, I have shown that
$$
\begin{equation}
\sum_{n = 1}^\infty \frac{H_n}{n^3} = \frac{5}{4}\cdot \frac{\pi^4}{90} = \frac{5}{4} \zeta(4). \tag{1}
\end{equation}
$$
Here
$$
H_n = \sum_{k = 1}^n \frac{1}{k}
$$
is the $n$th harmonic number. Here are a couple more examples. I will use the first to derive $(1)$.
*
*With the method from this answer, one can show that
$$
\int_0^{\pi/2} x^2\log^2(2\cos{x})\,dx = \frac{1}{5}\left(\frac{\pi}{2}\right)^5 + \pi \int_0^\infty y \log^2(1-e^{-2y})\,dy,
$$
and it then follows from the result of your question that
$$
\pi \int_0^\infty y \log^2(1- e^{-2y})\,dy = \frac{11}{45}\left(\frac{\pi}{2}\right)^5 - \frac{1}{5}\left(\frac{\pi}{2}\right)^5 = \frac{\pi}{8} \cdot \frac{\pi^4}{90}.
$$
*I also shared this example in the other answer, but for completeness, let me mention that
$$
\int_0^{\pi/2} x^2\log^2(2\cos{x})\,dx = \frac{1}{30}\left(\frac{\pi}{2}\right)^5 + \frac{1}{6}\int_0^{\pi/2} \log^4(2\cos{x})\,dx,
$$
hence
$$
\int_0^{\pi/2} \log^4(2\cos{x})\,dx = \frac{19}{15}\left(\frac{\pi}{2}\right)^5.
$$
More details about the derivations of 1. and 2. are contained in the answer I referenced above (near the bottom).
Let me now turn to deriving $(1)$. Taking $x = - e^{-2y}$ in the series expansion
$$
\log{(1+x)} = \sum_{n = 1}^\infty \frac{(-1)^{n+1}x^n}{n}
$$
and inserting the result into the integral evaluated in 1. gives
$$
\begin{align}
\frac{1}{8}\cdot\frac{\pi^4}{90} &= \int_0^\infty y \log^2(1 - e^{-2y})\,dy \\
& = \int_0^\infty y \left(\sum_{n = 1}^\infty \frac{e^{-2ny}}{n}\right)^2\,dy \\
& = \sum_{m = 1}^\infty\sum_{n = 1}^\infty\frac{1}{nm} \int_0^\infty y e^{-2(n + m)y}\,dy \\
& = \frac{1}{4}\sum_{m = 1}^\infty\sum_{n = 1}^\infty \frac{1}{nm(n+m)^2}. \tag{2}
\end{align}
$$
Now put $r = m+n$ and $s=n$ in order to write
$$
\begin{align}
\sum_{m = 1}^\infty\sum_{n = 1}^\infty \frac{1}{nm(n+m)^2} & = \sum_{r = 2}^\infty \frac{1}{r^2}\sum_{s = 1}^{r-1} \frac{1}{s(r-s)} = 2 \sum_{r=2}^\infty \frac{1}{r^3} \sum_{s=1}^{r-1}\frac{1}{s}, \tag{3}
\end{align}
$$
with the last equation a consequence of the identity $\frac{1}{s(r-s)} = \frac{1}{r}\left(\frac{1}{s}+ \frac{1}{r-s}\right)$. Insert $(3)$ into $(2)$ and multiply both sides by $2$ to get
$$
\frac{1}{4} \cdot \frac{\pi^4}{90} = \sum_{r=2}^\infty \frac{1}{r^3} \sum_{s=1}^{r-1}\frac{1}{s} = \sum_{r=2}^\infty \frac{H_{r-1}}{r^3}.
$$
Since $H_r - H_{r-1} = 1/r$, the identity $(1)$ now follows from the equations
$$
\frac{\pi^4}{90} = \sum_{r=1}^\infty \frac{1}{r^4} = \sum_{r = 1}^\infty \frac{H_r}{r^3} - \sum_{r=2}^\infty \frac{H_{r-1}}{r^3} = \sum_{r = 1}^\infty \frac{H_r}{r^3} - \frac{1}{4}\cdot\frac{\pi^4}{90}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/119253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"answer_id": 1
}
|
For complex $z$, find the roots $z^2 - 3z + (3 - i) = 0$
Find the roots of:
$z^2 - 3z + (3 - i) = 0$
$(x + iy)^2 - 3(x + iy) + (3 - i) = 0$
$(x^2 - y^2 - 3x + 3) + i(2xy -3y - 1) = 0$
So, both the real and imaginary parts should = 0. This is where I got stuck since there are two unknowns for each equation. How do I proceed?
|
$$9y^2+6y+1-4y^4-18y^2-6y+12y^2=0$$
$$-4y^2+3y^2+1=0`×(-1)$$
$$4y^2-3y^2-1=0$$
$$(4y^2+1)(y^2-1)=0$$
$$y^2=1$$
$$y=+1, x=(3+1)/2= 2$$
$$y=-1, x=(-3+1)/-2=1$$
$$z=x+yi$$
$$z1=2+i$$
$$z2=1-i$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/119626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what method to use.
|
You don't need to use induction here. Just reduce everything modulo 9:
$10 \equiv 1 \mod 9$, so $10^n \equiv 1 \mod 9$, so $10^n+5 \equiv 6 \mod 9$.
To find the possible values of $4^n \mod 9$, note that $4^3 \equiv 1 \mod 9$, so $4^n$ must be $1, 4,$ or $7 \mod 9$. Thus $3 \cdot 4^n$ is equal to $3 \mod 9$.
Hence $10^n + 3 \cdot 4^m + 5 \equiv 0 \mod 9$ for any positive integers $m, n$, not only when $n = m$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/120649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 5
}
|
How to compute $\sum\limits_{k=0}^n (-1)^k{2n-k\choose k}$? I got stuck at the computation of the sum
$$
\sum\limits_{k=0}^n (-1)^k{2n-k\choose k}.
$$
I think there is no purely combinatorial proof here since the sum can achieve negative values. Could you give me solution, it seems to involve generating functions.
|
Let
$$
a_n=\sum_{k=0}^n(-1)^k\binom{n-k}{k}
$$
Noting that
$$
\sum_{n=k}^\infty\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}\tag{1}
$$
we can compute the generating function of $a_n$:
$$
\newcommand{\cis}{\operatorname{cis}}
\begin{align}
\sum_{n=0}^\infty a_nx^n
&=\sum_{n=0}^\infty x^n\sum_{k=0}^n(-1)^k\binom{n-k}{k}\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^kx^n\binom{n-k}{k}\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^kx^{n+k}\binom{n}{k}\\
&=\sum_{k=0}^\infty(-1)^kx^k\frac{x^k}{(1-x)^{k+1}}\\
&=\frac{1}{1-x}\sum_{k=0}^\infty(-1)^k\left(\frac{x^2}{1-x}\right)^k\\
&=\frac{1}{1-x}\frac{1}{1+\frac{x^2}{1-x}}\\
&=\frac{1}{1-x+x^2}\\
&=\frac{1}{\alpha-\beta}\left(\frac{1}{x-\alpha}-\frac{1}{x-\beta}\right)\tag{2}
\end{align}
$$
where $\alpha$ and $\beta$ are the roots of $1-x+x^2=0$; $\alpha=\cis(\pi/3)$ and $\beta=\cis(-\pi/3)$.
Thus, the series for $(2)$ is
$$
\begin{align}
&\frac{1}{\cis(\pi/3)-\cis(-\pi/3)}\left(\frac{\cis(\pi/3)}{1-\cis(\pi/3)x}-\frac{\cis(-\pi/3)}{1-\cis(-\pi/3)x}\right)\\
&=\frac{2}{\sqrt{3}}\Im\left(\frac{\cis(\pi/3)}{1-\cis(\pi/3)x}\right)\\
&=\frac{2}{\sqrt{3}}\Im\left(\sum_{n=0}^\infty\cis((n+1)\pi/3)x^n\right)\tag{3}
\end{align}
$$
Therefore,
$$
a_n=\frac{2}{\sqrt{3}}\sin\left(\frac{n+1}{3}\pi\right)\tag{4}
$$
and the sequence requested above is
$$
a_{2n}=\frac{2}{\sqrt{3}}\sin\left(\frac{2n+1}{3}\pi\right)\tag{4}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/121407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 2
}
|
Using antiderivative to calculate complex integral $$\int_{1}^{3}(z-2)^3 dz $$
I get the following -
$$\frac{1}{4}[(3-2)^3 - (1-2)^3] = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$
However the answer sheet I have just show it reduced to $\frac{1}{4} - \frac{1}{4} = 0$
Cant see how they are getting a - instead of a +...what am I missing?
|
You should increase the power by 1 first to get:
$\frac{1}{4}[(3-2)^4 - (1-2)^4] = \frac{1}{4} - \frac{1}{4} = 0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/121846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.